52671 + Interview Questions in Chemistry in Class 11 Page 1018 InterviewSolution

50851.	A 1 M solution of glucose reaches dissociation equilibrium according to equation given below 6HCHOhArrC_(16)H_(12)O_(6). What is the concentration of HCHO at equilibrium constant is 6xx10^(22)
Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>.6xx10^(8)`M<br/>`3.2xx10^(6)`M<br/>`3.2xx10^(-4)`M<br/>`1.6xx10^(-4)`M</p>Solution :`6HCHOhArrC_(6)H_(<a href="https://interviewquestions.tuteehub.com/tag/12-269062" style="font-weight:bold;" target="_blank" title="Click to know more about 12">12</a>)O_(6)` forward <a href="https://interviewquestions.tuteehub.com/tag/reaction-22747" style="font-weight:bold;" target="_blank" title="Click to know more about REACTION">REACTION</a> <br/> `C_(6)H_(12)O_(6)hArr 6HCHO` backward reaction <br/> `K_(2) = [(1)/(K_(1))]^(1//6)` , `K_(2) = [(1)/(6xx10^(22))]^(1//6)` <br/>`K_(2) = 1.6xx10^(-4)` M</body></html>

Discussion

50852.	A 0.518 g sample of limestone is dissolved in HCl and then the calcium is precipitated as Ca_(2)C_(2)O_(4). After filtering and washing the precipitate, it requires 40 mL of 0.25 N KMnO_(4) solution acidified with H_(2)SO_(4) to titrate it as, MnO_(4)^(-) + H^(+) + C_(2)O_(4)^(2-) to CO_(2) + Mn^(2+) + 2H_(2)O The percentage of CaO in the sample is :
Answer» <html><body><p>0.54<br/>`27.1%`<br/>0.42<br/>0.84</p>Solution :Number of milliequivalents of `CaC_(2)O_(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>), KMnO_(4) " and" <a href="https://interviewquestions.tuteehub.com/tag/cao-411869" style="font-weight:bold;" target="_blank" title="Click to know more about CAO">CAO</a>` will be same. <br/> `40xx0.25=(W)/(56//2)xx1000` <br/> W=0.28 g (<a href="https://interviewquestions.tuteehub.com/tag/mass-1088425" style="font-weight:bold;" target="_blank" title="Click to know more about MASS">MASS</a> of CaO) <br/> `% CaO=(0.28)/(0.518)xx100=54%`</body></html>

Discussion

50853.	A 0.50 g H_(2)SO_(4) is dissolve in 0.25 L solution, so find out the normality and molarity respectively of the solution. (H=1, S= 32, O=16)
Answer» <html><body><p>0.040, 0.020<br/> 0.4, 0.2<br/>0.020, 0.040<br/> 0.2, 0.4</p>Solution :Molarity `=("Mass of <a href="https://interviewquestions.tuteehub.com/tag/solute-1217068" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTE">SOLUTE</a>(gm)")/("<a href="https://interviewquestions.tuteehub.com/tag/moleculer-2840274" style="font-weight:bold;" target="_blank" title="Click to know more about MOLECULER">MOLECULER</a> weight of solute" xx "<a href="https://interviewquestions.tuteehub.com/tag/volume-728707" style="font-weight:bold;" target="_blank" title="Click to know more about VOLUME">VOLUME</a> of solution(<a href="https://interviewquestions.tuteehub.com/tag/litre-1075864" style="font-weight:bold;" target="_blank" title="Click to know more about LITRE">LITRE</a>)")` <br/> `= (0.50)/(98xx0.5) = 0.020` <br/> `N= Mxx2 = 0.020xx2= 0.040`</body></html>

Discussion

50854.	A 0.518 g sample of lime stone is dissolved in HCl and then the calcium is precipitated as CaC_2O_4 After filtering and washing the precipitate, it requires 40.0 mL of 0.250 N KMnO_4solution acidified with H_2SO_4to titrate it as, MnO_(4)^(-)+H^(+) +C_(2)O_(4)^(2-) rarr Mn^(2+)+CO_(2)+2H_(2)O The percentage of CaO in the sample is:
Answer» <html><body><p>` 54.0% `<br/>`27.1% `<br/>` <a href="https://interviewquestions.tuteehub.com/tag/42-316129" style="font-weight:bold;" target="_blank" title="Click to know more about 42">42</a>%`<br/>` 84% `</p>Solution :No of <a href="https://interviewquestions.tuteehub.com/tag/milli-560824" style="font-weight:bold;" target="_blank" title="Click to know more about MILLI">MILLI</a> <a href="https://interviewquestions.tuteehub.com/tag/equivalents-974752" style="font-weight:bold;" target="_blank" title="Click to know more about EQUIVALENTS">EQUIVALENTS</a> of lime <a href="https://interviewquestions.tuteehub.com/tag/stone-1228007" style="font-weight:bold;" target="_blank" title="Click to know more about STONE">STONE</a> <br/> = no of milli equivalents of `CaC_(2)O_4` <br/> = noof milli equivalents of`KMnO_4` <br/> `= 40xx0.25 =10` <br/> No of milli moles of lime = 5 <br/>`:.` weight of `CaO = 5 xx 56 xx 10^(-3) = 0.28g` <br/>percentage of `CaO= (0.28)/(0.518)xx100=54.05%`</body></html>

Discussion

50855.	A 0.5 g sample containing MnO_(2) is treated with HCl liberating Cl_(2). The Cl_(2) is passed into a solution of KI and 30.0 cm^(3) of 0.1 M Na_(2)S_(2)O_(3) are required to tirate the liberated iodine. Calculate the percentage of MnO_(2) in the sample.
Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`<a href="https://interviewquestions.tuteehub.com/tag/30-304807" style="font-weight:bold;" target="_blank" title="Click to know more about 30">30</a>.0 mL 0.1 M Na_(2)S_(2)O_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)-=30.0 mL 0.1 N Na_(2)S_(2)O_(3)` <br/> `-=30.0 mL 0.1 N I_(2)` <br/> `-=30.0 mL 0.1 N Cl_(2)` <br/> `-=30.0 mL 0.1 N MnO_(2)` <br/> Amount of `MnO_(2)` present `=(NxxExxV)/(<a href="https://interviewquestions.tuteehub.com/tag/1000-265236" style="font-weight:bold;" target="_blank" title="Click to know more about 1000">1000</a>)` <br/> `=(1)/(10)xx(87)/(2)xx(30)/(1000)` <br/> `% MnO_(2)=(87xx30xx100)/(10xx2xx1000xx0.5)=26.1`</body></html>

Discussion

50856.	A 0.5 percent aqueous solution of KCl was found to freeze at 272.76K. Calculate the Van't Hoff factor and degree of dissociation of the solute at this concentration (K_(f)" for water = 1.86 k.kg.mol"^(-1)). Normal molar mass of KCl = 74.5.
Answer» <html><body><p></p>Solution :`T^(@)<a href="https://interviewquestions.tuteehub.com/tag/f-455800" style="font-weight:bold;" target="_blank" title="Click to know more about F">F</a>" of water "=273K` <br/> `T_(f)" of the solution "=272.76K` <br/> `DeltaT_(f)=T_(f^(@))-T_(f)=+0.24K` <br/> `M_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)=(K_(f)W_(2))/(DeltaT_(f).W_(1))` <br/> <a href="https://interviewquestions.tuteehub.com/tag/observed-582965" style="font-weight:bold;" target="_blank" title="Click to know more about OBSERVED">OBSERVED</a> molecular mass <br/> `M_(2)=(1.86xx0.5xx1000)/(100xx0.24)="38.75 g.mol"^(-1)` <br/> The colligative property is inversely related to the molar mass. <br/> `<a href="https://interviewquestions.tuteehub.com/tag/therefore-706901" style="font-weight:bold;" target="_blank" title="Click to know more about THEREFORE">THEREFORE</a>"Van.t Hoff factor"` <br/> `i=("Observed colligative property")/("Normal colligative property")` <br/> `=("Theoretical molar mass")/("Observed molar mass")` <br/> `"Vant Hoff factor (i) "=(74.5)/(38.75)=1.92` <br/> `"Degree of dissociation "alpha=(i-1)/(n-1)` <br/> `n="2 for <a href="https://interviewquestions.tuteehub.com/tag/kcl-527837" style="font-weight:bold;" target="_blank" title="Click to know more about KCL">KCL</a>"` <br/> `therefore alpha=(1.92-1)/(2-1)` <br/> `therefore"Degree of dissociation "=0.92`</body></html>

Discussion

50857.	A 0.5 percent aqueous solution of KCl was found to freeze at 272.76K. Calculate the Van.t Hoff factor and degree of dissociation of the solute at this concentration (K_(f)" for water = 1.86 k.kg.mol"^(-1)). Normal molar mass of KCl = 74.5.
Answer» <html><body><p></p>Solution :`T^(@)<a href="https://interviewquestions.tuteehub.com/tag/f-455800" style="font-weight:bold;" target="_blank" title="Click to know more about F">F</a>" of water "=273K` <br/> `T_(f)" of the solution "=272.76K` <br/> `DeltaT_(f)=T_(f^(@))-T_(f)=+0.24K` <br/> `M_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)=(K_(f)W_(2))/(DeltaT_(f).W_(1))` <br/> Observed molecular mass <br/> `M_(2)=(1.86xx0.5xx1000)/(100xx0.24)="38.75 g.mol"^(-1)` <br/> The colligative property is <a href="https://interviewquestions.tuteehub.com/tag/inversely-7688775" style="font-weight:bold;" target="_blank" title="Click to know more about INVERSELY">INVERSELY</a> related to the molar mass. <br/> `therefore"Van.t Hoff factor"` <br/> `i=("Observed colligative property")/("<a href="https://interviewquestions.tuteehub.com/tag/normal-1123860" style="font-weight:bold;" target="_blank" title="Click to know more about NORMAL">NORMAL</a> colligative property")` <br/> `=("Theoretical molar mass")/("Observed molar mass")` <br/> `"Vant Hoff factor (i) "=(74.5)/(38.75)=1.92` <br/> `"Degree of dissociation "alpha=(i-1)/(n-1)` <br/> `n="2 for <a href="https://interviewquestions.tuteehub.com/tag/kcl-527837" style="font-weight:bold;" target="_blank" title="Click to know more about KCL">KCL</a>"` <br/> `therefore alpha=(1.92-1)/(2-1)` <br/> `therefore"Degree of dissociation "=0.92`</body></html>

Discussion

50858.	A 0.5 g sample of an iron containing mineral mainly in the form of CuFeS_2 was reduced suitable to convert all the ferric ions into the ferrous form and was obtained as a solution. In the absence of any interfering matter, the solution required 42 " mL of " 0.1 M K_2Cr_2O_7 solution for titration calculate the percentage of CuFeS_2 in the mineral (Cu=63.5,Fe=55.8)
Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :Equivalent <a href="https://interviewquestions.tuteehub.com/tag/weight-1451304" style="font-weight:bold;" target="_blank" title="Click to know more about WEIGHT">WEIGHT</a> of `CuFeS_2=` molecular weight <br/> `CuFeS_2toCuSdarr+FeS` <br/> `FeS+H_2SO_4toFeSO_4+H_2Suarr` <br/> `2FeSO_4+H_2SO_4+OtoFe(SO_4)_3+H_2O` <br/> Molecular weight of `CuFeS_2=63.5+55.8+64=183.3g` <br/> `42 " <a href="https://interviewquestions.tuteehub.com/tag/ml-548251" style="font-weight:bold;" target="_blank" title="Click to know more about ML">ML</a> of " 0.01 M K_2Cr_2O_7=42xx0.01xx6NK_2Cr_2O_7` <br/> `=<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>.52 m" Eq of "K_2Cr_2O_7` <br/> `=2.52m" Eq of "<a href="https://interviewquestions.tuteehub.com/tag/fe-460008" style="font-weight:bold;" target="_blank" title="Click to know more about FE">FE</a>^(2+)` <br/> `=2.52 m" Eq of "CuFeS_2` <br/> `=2.52 m" Eq of "CuFeS_2` <br/> `=2.52xx10^(-3)xx18.3g of CuFeS_2` <br/> `=0.4619 g of CuFeS_2` <br/> `%` of `CuFeS_2=(0.4619xx100)/(0.5)=92.38%`</body></html>

Discussion

50859.	(a) 0.456 g of a metal gives 0.606 g of its chloride. Calculate the equivalent mass of the metal (b)Calculate the equivalent mass of potassium dichromate. The reduction half-reaction in acid medium is,Cr_(2)O_(7)^(2-)+14H^(+)+6e^(-) rarr 2Cr^(3+)+7H_(2)O
Answer» <html><body><p></p>Solution :Mass of the <a href="https://interviewquestions.tuteehub.com/tag/metal-1094457" style="font-weight:bold;" target="_blank" title="Click to know more about METAL">METAL</a> = `W_(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)` = 0.456 g.<br/>Mass of the metal chloride = `W_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)` = 0.606 g<br/>`:.` Mass of chlorine = `W_(2) -W_(1)` = 0.606 - 0.456 = 0.15 g<br/>0.15 g of chlorine combine with 0.456 g of metal. <br/>`:.` 35.46 g of chlorine will combine with`0.456/0.15 xx 35.46` = 107.79 g `<a href="https://interviewquestions.tuteehub.com/tag/eq-446394" style="font-weight:bold;" target="_blank" title="Click to know more about EQ">EQ</a> ^(-1)`. <br/>`Cr_(2)O_(7)^(2-)+14H^(+)+6e^(-) rarr 2Cr^(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>+)+7H_(2)O`<br/>Equivalent mass = `("Molar mass")/("No. of electrons gained")`<br/>`:.` Molar mass of `K_(2)Cr_(2)O_(7)` = 294.18<br/>`:.` Equivalent mass of `K_(2)Cr_(2)O_(7) = 294.18/6 = 49.03`</body></html>

Discussion

50860.	A 0.25 M gulcose solution at 370.28 K has approximately the pressure as blood does what is the osmotic pressure of blood?
Answer» <html><body><p></p>Solution :C=0.25 M <br/>T=370.28 <a href="https://interviewquestions.tuteehub.com/tag/k-527196" style="font-weight:bold;" target="_blank" title="Click to know more about K">K</a> <br/> `(<a href="https://interviewquestions.tuteehub.com/tag/pi-600185" style="font-weight:bold;" target="_blank" title="Click to know more about PI">PI</a>)_("gulcose")`=CRT <br/>`(pi)=0.25 mol <a href="https://interviewquestions.tuteehub.com/tag/l-535906" style="font-weight:bold;" target="_blank" title="Click to know more about L">L</a>^(-<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)xx0.082 L atm^(-1)mol^(-1) xx370.28K` <brge7.59 atm=""></brge7.59></body></html>

Discussion

50861.	A 0.239g sample of a gas in a 100-ml flask exerts a pressure of 600mm Hg at 14^(@)C. What is the gas?
Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/chlorine-915874" style="font-weight:bold;" target="_blank" title="Click to know more about CHLORINE">CHLORINE</a> <br/>Nitrogen<br/>Krypton<br/>Xenon</p>Answer :A</body></html>

Discussion

50862.	A 0.2 g sample of benzoic acid, C_(6)H_(5)COOH, is titrated with a 0.120 M Ba(OH)_(2) solution. What volume of the Ba(OH)_(2) solution is required to reach the equivalence point ? ("Substance")/(C_(6)H_(5)COOH) "" ("Molar mass")/(122.1 g " mol"^(-1))
Answer» <html><body><p>6.82 mL<br/>13.6 mL<br/>17.6 mL<br/>35.2 mL</p>Solution :N//A</body></html>

Discussion

50863.	A 0.141g sample of phosphorus containing compound was digested in a mixture ofHNO_(3) and H_(2)SO_(4) which resulant in formnation of CO_(2), H_(2)O and H_(3)PO_(4). Addition of ammounium molybdate yielded a solid having the composition (NH_(4))_(3) PO_(4).12MoO_(3). The precipitate was filtered, washed and dissolved in 50.0mL of 0.20M NaOH. (NH_(4))_(3) PIO_(4).12MoO_(3(s)) + 26OH^(-) rarr HPO_(4)^(2-) + 12MoO_(4)^(2-) + 14H_(2)O + 3NH_(3(g)) After boilingthe solution to remove the NH_(3), the excess of NaOH was titrated with 14.1 mL of 0.174M, HCl. Calculate the percentage of phosprous in the sample.
Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :`6.4%`</body></html>

Discussion

50864.	A 0.1g sample of pyrolusite ore containing MnO_(2)(MW = 87) was dissolved in a concentrated HCI solutiona and liberated CI_(2)(g) was passed through a concentrated KI solution releasing I_(2) if the liberated iodine required 16 ml of 1.25M sodium thiosulphate (Na_(2)S_(2)O_(3)) solution, mass percentage of MnO_(2) in the given sample is
Answer» <html><body><p>`43.5%`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/87-339709" style="font-weight:bold;" target="_blank" title="Click to know more about 87">87</a>%`<br/>`21.75%`<br/>`0.875%`</p>Solution :No. of m.eqts `MnO_(2) =` No. of m.eqts of `Cl_(2)` <br/> = No. of m.eqts of `<a href="https://interviewquestions.tuteehub.com/tag/ki-533104" style="font-weight:bold;" target="_blank" title="Click to know more about KI">KI</a> =` No. of m.eqts hypo No. of m.eqts <br/> `= MnO_(2) = <a href="https://interviewquestions.tuteehub.com/tag/16-276476" style="font-weight:bold;" target="_blank" title="Click to know more about 16">16</a> xx 1.25 = 16 xx (5)/(4) =20` <br/> No. of milli <a href="https://interviewquestions.tuteehub.com/tag/moles-1100459" style="font-weight:bold;" target="_blank" title="Click to know more about MOLES">MOLES</a> of `MnO_(2) = 10` <br/> <a href="https://interviewquestions.tuteehub.com/tag/weight-1451304" style="font-weight:bold;" target="_blank" title="Click to know more about WEIGHT">WEIGHT</a> of `MnO_(2) = 10 xx 87 xx 10^(-3) = 0.87g` <br/> weight % of `MnO_(2) = (0.87)/(1) xx 100 = 87%`</body></html>

Discussion

50865.	A 0.1 M solution of weak acid HA is 1% dissociated at 25^(@)C To this solution NaA is added till [NaA] = 0.2 M, if the new degree of dissociation of HA = yxx 10^(-5) then what is 'y'?
Answer» <html><body><p><br/></p>Solution :` alpha = 0.01 , C =0.1` <br/> ` K_a = <a href="https://interviewquestions.tuteehub.com/tag/calpha-2020052" style="font-weight:bold;" target="_blank" title="Click to know more about CALPHA">CALPHA</a> ^(2)= 0.1 xx ( 0.01 )^(2)= <a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a> ^(5) ` <br/> ` K_a = ([<a href="https://interviewquestions.tuteehub.com/tag/h-1014193" style="font-weight:bold;" target="_blank" title="Click to know more about H">H</a>^(+) ] [A^(-) ]) /( [<a href="https://interviewquestions.tuteehub.com/tag/ha-479237" style="font-weight:bold;" target="_blank" title="Click to know more about HA">HA</a>]) =([H^(+) ]xx 0.2 )/(0.1 )` <br/> ` <a href="https://interviewquestions.tuteehub.com/tag/rarr-1175461" style="font-weight:bold;" target="_blank" title="Click to know more about RARR">RARR</a> [H^(+) ] = calpha =10 ^(-5)xx ( 0.1)/( 0.2 ) rArr alpha = 5 xx 10 ^(-5)`</body></html>

Discussion

50866.	A 0.05 N solution of acetic acid is found to be 1.9% ionized at 25^(@)C. Calculate (i) K_(a) for acetic acid and(ii) the pH of the solution.
Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :Amount of <a href="https://interviewquestions.tuteehub.com/tag/acetic-1967065" style="font-weight:bold;" target="_blank" title="Click to know more about ACETIC">ACETIC</a> <a href="https://interviewquestions.tuteehub.com/tag/acid-847491" style="font-weight:bold;" target="_blank" title="Click to know more about ACID">ACID</a> dissociated `=0.5xx1.9//100=0.00095` <br/> `:. ` Concs. At equilibrium : `[CH_(3)<a href="https://interviewquestions.tuteehub.com/tag/co-920124" style="font-weight:bold;" target="_blank" title="Click to know more about CO">CO</a> OH]=0.05-0.00095 M, [CH_(3)CO O^(-)] = [H^(+)] = 0.00095 M`</body></html>

Discussion

50867.	A 0.02 M solution of pyridinium hydrochloride has pH=3.44 . Calculate the ionization constantof pyridine.
Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :Pyridinium hydrochloride `C_6H_5N^(+) <a href="https://interviewquestions.tuteehub.com/tag/hcl-479502" style="font-weight:bold;" target="_blank" title="Click to know more about HCL">HCL</a>^(-)` is conjugate <a href="https://interviewquestions.tuteehub.com/tag/acid-847491" style="font-weight:bold;" target="_blank" title="Click to know more about ACID">ACID</a> of pyridine `(C_6H_5N)`. <br/> pH=3.44 = -log `[H^+]` <br/> `therefore` log `[H^+] = -3.44 = bar4.56` <br/> `therefore [H^+] = 10^(-3.44) = 3.63xx10^(-4) = alpha` <br/> `{:(,C_6H_5N^(+)HCl^(-)+ aqhArr ,C_6H_5NCl_((aq))^(-) +, H_((aq))^(+)),("Molarity at equilibrium" , (0.012-alpha)M =0.02 M, 3.63xx10^(-4) , 3.63xx10^(-4)) :}` <br/> `K_a=([C_6H_5NCl^(-)][H^+])/([C_6H_5N^(+)HCl^-])` <br/> `=((3.63xx10^(-4))(3.63xx10^(-4)))/0.02=6.588xx10^(-6)` <br/> Pyridine is conjugate base of pyridinium hydrochloride , so its ionization constant = `K_b` <br/> `K_b=K_w/K_a=(1.0xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/14-272882" style="font-weight:bold;" target="_blank" title="Click to know more about 14">14</a>))/(6.588xx10^(-6))=1.5179xx10^(-7)`</body></html>

Discussion

50868.	A 0.02 M solution of pyridinium hydrochloride has pH=3.44. Calculate the ionization constant of pyridine.
Answer» <html><body><p></p>Solution :`pH = 3.44 , i.e., log [H^(+)]=-3.44 = bar(4).56 :.[H^(+)]=3.63xx10^(_4)M` <br/> `C_(5)H_(5)N.HCl+aq hArr C_(5)H_(5)overset(+)<a href="https://interviewquestions.tuteehub.com/tag/ncl-570043" style="font-weight:bold;" target="_blank" title="Click to know more about NCL">NCL</a>^(-)+H^(+)` <br/> `K_(a)=([C_(5)H_(5)overset(+)NCl^(-)][H^(+)])/([C_(5)H_(5)N.HCl])=((3.63xx10^(-4))(3.63xx10^(-4)))/(2xx10^(-2))=6.588xx10^(-6)` <br/> `pK_(a)=-log(6.588xx10^(-6))=6-0.8187=5.<a href="https://interviewquestions.tuteehub.com/tag/18-278913" style="font-weight:bold;" target="_blank" title="Click to know more about 18">18</a>` <br/> `pK_(a)+pK_(b)=14 :. pK_(b)=14-5.18 = 8.82` <br/> `-log K_(b) = 8.82 or log K_(b) = - 8.82 = bar(-).18 :. K_(b)=1.514xx10^(-9)` <br/> Alternatively, Pyridine hydrochloride is a salt of weak <a href="https://interviewquestions.tuteehub.com/tag/base-892693" style="font-weight:bold;" target="_blank" title="Click to know more about BASE">BASE</a> and strong acid . <a href="https://interviewquestions.tuteehub.com/tag/hence-484344" style="font-weight:bold;" target="_blank" title="Click to know more about HENCE">HENCE</a>, <br/> `pH=-(1)/(2)[log K_(w)-logK_(b)+logc],i.e., 3.44 = - (1)/(2) [-14-logK_(b)+log(2xx10^(-2))]` <br/> or `6.88=14+log K_(b) + 1.70 or log K_(b) = - 8.82 = bar(9).18 or K_(b)=1.5xx10^(-9)`.</body></html>

Discussion

50869.	A 0.01 M solution of glucose in water freezes at -0.0186^(@)C A 0.01 M solution of KNO_(3) in water will freeze at
Answer» <html><body><p>`-0.0093^(@)C`<br/>`-0.0372^(@)C`<br/>`-0.0186^(@)C`<br/>`-0.093^(@)C`</p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`KNO_(3)` ionises as, <br/> `KNO_(3) <a href="https://interviewquestions.tuteehub.com/tag/harr-2692945" style="font-weight:bold;" target="_blank" title="Click to know more about HARR">HARR</a> K^(+) + NO_(3)^(-)` <br/> `therefore` Concentration in solution `=2xx0.01=0.02` <br/> `DeltaT_(f)=T_(f)-T^(@)=K_(f)xxm`for glucose <br/> `therefore k_(f)(<a href="https://interviewquestions.tuteehub.com/tag/deltatf-2053493" style="font-weight:bold;" target="_blank" title="Click to know more about DELTATF">DELTATF</a>)/(m)=(0.018)/(0.01)=1.86^(@) m^(-1)` <br/> `DeltaT_(f)` for `KNO_(3)=1.86xx0.02=0.0372^(@)C` <br/> `T_(f)=T^(@)-T_(f)=0.0000-0.0372=-0.0372^(@)C`</body></html>

Discussion

50870.	A 0.01 M solution of acetic acid is 1.34 % ionised (degree of dissociation = 0.0134) at 298 K. What is theionization constant of acetic acid ?
Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`{:(,CH_(3)CO OH,hArr,CH_(3)CO O^(-) , +,<a href="https://interviewquestions.tuteehub.com/tag/h-1014193" style="font-weight:bold;" target="_blank" title="Click to know more about H">H</a>^(+),,),("Initial <a href="https://interviewquestions.tuteehub.com/tag/conc-927968" style="font-weight:bold;" target="_blank" title="Click to know more about CONC">CONC</a>.",C "mol" <a href="https://interviewquestions.tuteehub.com/tag/l-535906" style="font-weight:bold;" target="_blank" title="Click to know more about L">L</a>^(-1),,0,,0,,),("At. eqm.",c(1-alpha),, C alpha,,C alpha,,):}` <br/> `<a href="https://interviewquestions.tuteehub.com/tag/k-527196" style="font-weight:bold;" target="_blank" title="Click to know more about K">K</a>=(C alpha.C alpha)/(C (1-alpha))=(C alpha^(2))/(1-alpha)=(0.01xx(0.0134)^(2))/(1-00.01)=1.8xx10^(-6)`</body></html>

Discussion

50871.	A 100 watt bulb emits monochromatic light of wavelength 400 nm . Calculate the number of photons emitted per second by the bulb.
Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :`<a href="https://interviewquestions.tuteehub.com/tag/2012-290617" style="font-weight:bold;" target="_blank" title="Click to know more about 2012">2012</a> <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> <a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>^(<a href="https://interviewquestions.tuteehub.com/tag/20-287209" style="font-weight:bold;" target="_blank" title="Click to know more about 20">20</a>)`</body></html>

Discussion

50872.	9.9g of amide with molecular formula C_(4)H_(5)N_(x)O_(y) on heating with alkali liberated 1.7g of ammonia. If the percentage of oxygen is 33.33% then the ratio of 'N' and 'O' atoms in the compound is:
Answer» <html><body><p>`1:1`<br/>`1:2`<br/>`2:3`<br/>`3:2`</p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :Amide `C_(4)H_(5)N_(x)O_(y)` will give x mol `NH_(3)`. <br/> `:.` <a href="https://interviewquestions.tuteehub.com/tag/molecular-562994" style="font-weight:bold;" target="_blank" title="Click to know more about MOLECULAR">MOLECULAR</a> <a href="https://interviewquestions.tuteehub.com/tag/mass-1088425" style="font-weight:bold;" target="_blank" title="Click to know more about MASS">MASS</a> of amide `=(9.9)/(1.7) xx 17x = 99x` <br/> % <a href="https://interviewquestions.tuteehub.com/tag/nitrogen-1118291" style="font-weight:bold;" target="_blank" title="Click to know more about NITROGEN">NITROGEN</a> in the amide `= (14x)/(99x) xx 100 = <a href="https://interviewquestions.tuteehub.com/tag/14-272882" style="font-weight:bold;" target="_blank" title="Click to know more about 14">14</a>.14` <br/> Ratio of number of atoms of 'N' and 'O' <br/> `= (14.14)/(14): (33.33)/(16) = 1:2`</body></html>

Discussion

50873.	99.95% dihydrogen is obtained by electrolysing warm aqueous Ba(OH)_2 solution by electrodes……..
Answer» <html><body><p>Cu<br/>Pt<br/>Ni<br/>Zn</p>Answer :<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a></body></html>

Discussion

50874.	9.8g of an acid of molecuar weight 98 was neutralised by 200ml of one normal caustic soda. What is thebasicity of the acid?
Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/number-582134" style="font-weight:bold;" target="_blank" title="Click to know more about NUMBER">NUMBER</a> of milli <a href="https://interviewquestions.tuteehub.com/tag/equivalent-446407" style="font-weight:bold;" target="_blank" title="Click to know more about EQUIVALENT">EQUIVALENT</a> of <a href="https://interviewquestions.tuteehub.com/tag/base-892693" style="font-weight:bold;" target="_blank" title="Click to know more about BASE">BASE</a> `=200xx1=200meq`. <br/> Number of milli equivalent of acid =200m. <a href="https://interviewquestions.tuteehub.com/tag/eq-446394" style="font-weight:bold;" target="_blank" title="Click to know more about EQ">EQ</a> <br/> 200 milli equicalent of acid =? <br/> Equivalent weight of acid `=(9.8xx1000)/(200)=<a href="https://interviewquestions.tuteehub.com/tag/49-318440" style="font-weight:bold;" target="_blank" title="Click to know more about 49">49</a>` <br/> Basicity of acid `=("Molecular weight ")/("Equivalent weight")=(98)/(49)=2`</body></html>

Discussion

50875.	"9.84 cm"^(3) of ethyl alcohol is withdrawn from a flask containing "45.3 cm"^(3) of the sample. What is the volume ethyl alcohol left in the flask ?
Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :<a href="https://interviewquestions.tuteehub.com/tag/subtract-1231765" style="font-weight:bold;" target="_blank" title="Click to know more about SUBTRACT">SUBTRACT</a> : `{:(45.2),(9.84),(<a href="https://interviewquestions.tuteehub.com/tag/bar-892478" style="font-weight:bold;" target="_blank" title="Click to know more about BAR">BAR</a>(35.36)):}` <br/> or 35.4 (after <a href="https://interviewquestions.tuteehub.com/tag/rounding-1191583" style="font-weight:bold;" target="_blank" title="Click to know more about ROUNDING">ROUNDING</a> off)</body></html>

Discussion

50876.	""(92)^(233)U is assumed to decay by emitting alpha- and beta-particles, thepossible produce of decay is
Answer» <html><body><p>`""_(82)^(207)Pb`<br/>`""_(82)^(<a href="https://interviewquestions.tuteehub.com/tag/205-292836" style="font-weight:bold;" target="_blank" title="Click to know more about 205">205</a>)Pb`<br/>`""_(82)^(206)Pb`<br/>`""_(82)^(208)Pb`</p>Solution :`(238)/(4)=59(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)/(2)` and `(206)/(4)=51(1)/(2)` <br/> `implies ""_(92)U^(238)` is the starting element of 4n+2 series <br/> `implies ""_(82)Pb^(206)` is the <a href="https://interviewquestions.tuteehub.com/tag/possible-592355" style="font-weight:bold;" target="_blank" title="Click to know more about POSSIBLE">POSSIBLE</a> <a href="https://interviewquestions.tuteehub.com/tag/decay-945588" style="font-weight:bold;" target="_blank" title="Click to know more about DECAY">DECAY</a> <a href="https://interviewquestions.tuteehub.com/tag/product-25523" style="font-weight:bold;" target="_blank" title="Click to know more about PRODUCT">PRODUCT</a> of this series.</body></html>

Discussion

50877.	Sodium carbonate of 92% purity is used in the reation Na_(2)CO_(3)+CaCl_(2)toCaCO_(3)+2NaCl. The number of grams of Na_(2)CO_(3) required to yield 1 gm of CaCO_(3)
Answer» <html><body><p>8.5g<br/>10.5g<br/>11.52g<br/>1.152g</p>Answer :D</body></html>

Discussion

50878.	9.2 grams of N_(2)O_(4) (g) is taken in a closed one litre vessel and heated till the following equilibrium is reachedN_(2) O_(4) (g) hArr 2 NO_(2) (g) At equilibrium , 50% N_(2)O_(4) (g) is dissociated . What is the equilibrium constant ( in mol lit^(-1)) (" Molecular weughtof" N_(2)O_(4) = 92)
Answer» <html><body><p>`0.1`<br/>`0.<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>`<br/>`0.4`<br/>2</p>Solution :Intial` [N_(2)O_(4)] = 9.2//92 " mol"<a href="https://interviewquestions.tuteehub.com/tag/l-535906" style="font-weight:bold;" target="_blank" title="Click to know more about L">L</a>^(-1) = 0.1 " mol "L^(-1)` <br/> At. Eqm. ( after 50% dissociation ), <br/> `[N_(2)O_(4)] = 0.05 M, [NO_(2)] = 0.1 M` <br/> ` <a href="https://interviewquestions.tuteehub.com/tag/k-527196" style="font-weight:bold;" target="_blank" title="Click to know more about K">K</a> = ([NO_(2)])/([N_(2)O_(4)])=(0.1)^(2)/0.05 = 0.2`</body></html>

Discussion

50879.	9.2 g of N_(2) O_(4 (g)) is taken in 1 lit vessel and heated . At equilibrium , 50 % is dissociated . Equilibrium constant (mol/lit) [MW = 92]
Answer» <html><body><p>0.1<br/>0.2<br/>0.4<br/>2</p>Answer :<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a></body></html>

Discussion

50880.	9.0g m ice at 0^(@)C is mixed with 36gm of water at 50^(@)C in a thermally insulated container, using the following data, answer the question that follow: C_(P (H_(2)O)) = 4.18 Jg^(-1) K^(-1), Delta H_("fusion") (ice) = 335 J g^(-1) Delta S_("ice") is
Answer» <html><body><p>`11.04 <a href="https://interviewquestions.tuteehub.com/tag/jk-521675" style="font-weight:bold;" target="_blank" title="Click to know more about JK">JK</a>^(-1)`<br/>`3.16 JK^(-1)`<br/>`14.<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a> JK^(-1)`<br/>`7.84 JK^(-1)`</p>Solution :`Delta S_(ice) = (Delta H_("<a href="https://interviewquestions.tuteehub.com/tag/fusion-1002691" style="font-weight:bold;" target="_blank" title="Click to know more about FUSION">FUSION</a>"))/(T_((<a href="https://interviewquestions.tuteehub.com/tag/mp-565167" style="font-weight:bold;" target="_blank" title="Click to know more about MP">MP</a>))) + w underset(T_(1))overset(T_(2))int (Cp.dT)/(T)` <br/> `= (9 xx 335)/(273) + 9 xx 4.18 "<a href="https://interviewquestions.tuteehub.com/tag/ln-1076444" style="font-weight:bold;" target="_blank" title="Click to know more about LN">LN</a>" (296.97)/(273) = 14.206`</body></html>

Discussion

50881.	90g Na and 170g K is present in a person with 70 kg of body weight.
Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :<a href="https://interviewquestions.tuteehub.com/tag/true-713260" style="font-weight:bold;" target="_blank" title="Click to know more about TRUE">TRUE</a> <a href="https://interviewquestions.tuteehub.com/tag/statement-16478" style="font-weight:bold;" target="_blank" title="Click to know more about STATEMENT">STATEMENT</a></body></html>

Discussion

50882.	9.0g m ice at 0^(@)C is mixed with 36gm of water at 50^(@)C in a thermally insulated container, using the following data, answer the question that follow: C_(P (H_(2)O)) = 4.18 Jg^(-1) K^(-1), Delta H_("fusion") (ice) = 335 J g^(-1) Final tempeature of water is
Answer» <html><body><p>304.43K<br/>296.97 K<br/>303.93K<br/>287K</p>Solution :If final temperature is T then we can <a href="https://interviewquestions.tuteehub.com/tag/give-468520" style="font-weight:bold;" target="_blank" title="Click to know more about GIVE">GIVE</a> expression as <a href="https://interviewquestions.tuteehub.com/tag/like-1073845" style="font-weight:bold;" target="_blank" title="Click to know more about LIKE">LIKE</a> this `36 xx 4.18 xx (<a href="https://interviewquestions.tuteehub.com/tag/323-307610" style="font-weight:bold;" target="_blank" title="Click to know more about 323">323</a>- T)` <br/> `=9 xx 335 + 9 xx (T - 273) xx 4.18` <br/> `48605 - 150.48T = 3015 + 37.62T - 10270.26` <br/> On <a href="https://interviewquestions.tuteehub.com/tag/solving-1217196" style="font-weight:bold;" target="_blank" title="Click to know more about SOLVING">SOLVING</a>, T = 296.97K</body></html>

Discussion

50883.	9.0g m ice at 0^(@)C is mixed with 36gm of water at 50^(@)C in a thermally insulated container, using the following data, answer the question that follow: C_(P (H_(2)O)) = 4.18 Jg^(-1) K^(-1), Delta H_("fusion") (ice) = 335 J g^(-1) Delta S_("water") is
Answer» <html><body><p>`-12.64 JK^(-1)`<br/>`-0.34 JK^(-1)`<br/>`-5.42 JK^(-1)`<br/>`12.64 JK^(-1)`</p>Solution :`<a href="https://interviewquestions.tuteehub.com/tag/delta-947703" style="font-weight:bold;" target="_blank" title="Click to know more about DELTA">DELTA</a> S_("<a href="https://interviewquestions.tuteehub.com/tag/water-1449333" style="font-weight:bold;" target="_blank" title="Click to know more about WATER">WATER</a>") = w underset(T_(1))overset(T_(2))<a href="https://interviewquestions.tuteehub.com/tag/int-11513" style="font-weight:bold;" target="_blank" title="Click to know more about INT">INT</a> (<a href="https://interviewquestions.tuteehub.com/tag/cp-427206" style="font-weight:bold;" target="_blank" title="Click to know more about CP">CP</a> dT)/(T)` <br/> `=36 xx 4.18 "ln " (296.97)/(323) = - 12.643 JK^(-1)`</body></html>

Discussion

50884.	90.8 litres of a mixture of nitrogen and hydrogen measured at STP were passed over a catalyst. After the reaction, the volume of the mixture reduced to 68.1 litres. Ammonia thus formed was dissolved in 101 ml of an aqueous ammonia solution of density of 0.85 g//ml containing 12% by mass of NH_(4)OH. Determine the percent weight strength of the final solution. [Give answer exculding decimal places]
Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :<a href="https://interviewquestions.tuteehub.com/tag/44-316683" style="font-weight:bold;" target="_blank" title="Click to know more about 44">44</a></body></html>

Discussion

50885.	9.00 litres of a gas at 16 atm and 27°C weigh 93.6 g. What is the molecular mass of the gas ?
Answer» <html><body><p></p>Solution :In the present <a href="https://interviewquestions.tuteehub.com/tag/case-910082" style="font-weight:bold;" target="_blank" title="Click to know more about CASE">CASE</a>, <br/> `P=16 atm, V = 9.00 <a href="https://interviewquestions.tuteehub.com/tag/l-535906" style="font-weight:bold;" target="_blank" title="Click to know more about L">L</a>, T = 27^(@) C = 300 K, w = 93.6 g` <br/> According to the gas <a href="https://interviewquestions.tuteehub.com/tag/equation-974081" style="font-weight:bold;" target="_blank" title="Click to know more about EQUATION">EQUATION</a>, `<a href="https://interviewquestions.tuteehub.com/tag/pv-593601" style="font-weight:bold;" target="_blank" title="Click to know more about PV">PV</a> = w/M RT` <br/> or `M = (wRT)/(PV)` <br/> <a href="https://interviewquestions.tuteehub.com/tag/substituting-1231652" style="font-weight:bold;" target="_blank" title="Click to know more about SUBSTITUTING">SUBSTITUTING</a> the values, we get <br/> `M = (93.6 xx 0.0821 xx 300)/(16 xx 9.00) = 16.01`</body></html>

Discussion

50886.	90 g of acetic acid react with excess of NaHCO_(3) then what volume of CO_(2) will produce at S.T.P. Write your answer in terms of nearest integer.
Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :<a href="https://interviewquestions.tuteehub.com/tag/34-308171" style="font-weight:bold;" target="_blank" title="Click to know more about 34">34</a></body></html>

Discussion

50887.	896 mL vapour of a hydrocarbon 'A' having carbon 87.80%and hydrogen 12.19 % weighs 3.29 g at STP, Hydrogenation of 'A' gives 2-methylpentane. Also 'A' on hydration in the presence of H_2SO_4 and HgSO_4 gives a ketone 'B' having molecular formula C_6H_12O. The ketone 'B' gives a positive iodoform test. Find the structures of 'A' and give the reactions involved.
Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/step-25533" style="font-weight:bold;" target="_blank" title="Click to know more about STEP">STEP</a> 1 . To determine the molecular mass of hydrocarbon (A). <br/> 896 mL vapours of hydrocarbon (A) weigh at STP =3.28 g <br/> `therefore` 22700 mL vapours of A will weigh at STP `=(3.28xx22700)/896 "g mol"^(-1)=83.1 g` <br/> `therefore` Molecular mass of hydrocarbon (A) = `83.1 g mol^(-1)` <br/> Step 2. To determine the empirical formula of hydrocarbon (A). <br/> `{:("<a href="https://interviewquestions.tuteehub.com/tag/element-969236" style="font-weight:bold;" target="_blank" title="Click to know more about ELEMENT">ELEMENT</a>","% age","Atomic mass","<a href="https://interviewquestions.tuteehub.com/tag/relative-1183377" style="font-weight:bold;" target="_blank" title="Click to know more about RELATIVE">RELATIVE</a> ratio","Relative no. of atoms","Simplest ratio"),(C,87.8,12,7.31,1,3),(H,12.19,1,12.19,1.66,5):}` <br/> Thus, empirical formula of hydrocarbon (A) =`C_3H_5` and empirical formula mass = 12 x 3 + 5 x 1 = 41 u <br/> `therefore n="Molecular mass "/"Empirical formula mass "=83.1/41=2.02 approx 2` <br/> Thus, molecular formula of hydrocarbon (A)= 2 x Empirical formula = `2 xx C_3H_5 = C_6H_10` <br/> Step 3.To determine the structures of compounds (A) and (B) <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/PR_CHE_02_XI_C13_E14_002_S01.png" width="80%"/> <br/> (i)Since hydrogenation of hydrocarbon (A) requires 2 <a href="https://interviewquestions.tuteehub.com/tag/moles-1100459" style="font-weight:bold;" target="_blank" title="Click to know more about MOLES">MOLES</a> of hydrogen to form 2-methylpentane, therefore, hydrocarbon (A) is an alkyne having five carbon atoms in a straigth chain and a methyl substitutent at <a href="https://interviewquestions.tuteehub.com/tag/position-1159826" style="font-weight:bold;" target="_blank" title="Click to know more about POSITION">POSITION</a> 2. Thus, the two possible structures for the alkyne (A) are I and II : <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/PR_CHE_02_XI_C13_E14_002_S02.png" width="80%"/> <br/> (ii)Since addition of `H_2O` to alkyne (A) in presence of `Hg^(2+)` , gives a ketone which gives positive iodoform test, therefore, ketone (B) must be a methyl ketone , i.e., it must contain a `COCH_3` group. Now addition of `H_2O` to alkyne (II) should give a mixture of two ketone in which ketone (B) (Which shows +ve iodoform test ) predominates. <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/PR_CHE_02_XI_C13_E14_002_S03.png" width="80%"/> <br/> In contrast, addition of `H_2O` to alkyne (I) will give only one ketone , i.e., 4-methylpentan-2-one which gives iodoform test. <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/PR_CHE_02_XI_C13_E14_002_S04.png" width="80%"/> <br/>Thus, hydrocarbon (A) is 4-methylpent-1-yne.</body></html>

Discussion

50888.	896 mL vapour of a hydrocarbon 'A' having carbon 87.80% and hydrogen 12.19% weighs 3.28 gat STP. Hydrogenation of 'A' gives 2-methyl pentane. Also 'A' on hydration in the presence of H_(2)SO_(4) and HgSO_(4) gives a ketone 'B' moelcular formula C_(6)H_(12)O. The ketone 'B' gives a positive iodoformtest. Find the structure of 'A' and give the reactions involved.
Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :To calculate molar <a href="https://interviewquestions.tuteehub.com/tag/mass-1088425" style="font-weight:bold;" target="_blank" title="Click to know more about MASS">MASS</a> of hydrocarbon .A. it is given that 896 ml of Hydrocarbon .A. `C_(<a href="https://interviewquestions.tuteehub.com/tag/x-746616" style="font-weight:bold;" target="_blank" title="Click to know more about X">X</a>)H_(y)` Weighs 3.28 gat STP. <br/> 22400 ml of has `AC_(x)H_(y)` mass `= 3.28 xx 22400` ml/896ml = 831 g/<a href="https://interviewquestions.tuteehub.com/tag/mol-1100196" style="font-weight:bold;" target="_blank" title="Click to know more about MOL">MOL</a>. Hence, the molar mas of A = 831 g mol. <br/> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/KPK_AIO_CHE_XI_P2_C13_E04_049_S01.png" width="80%"/> <br/> Thus, empirical formula `=C_(3)H_(5)` <br/> Empirical formula weight `= 3 xx 12= 36 + 5 = 41`. <br/> Molecular formula = (Empirical formula)n<br/> n = Mol. mass/Empirical mass = 831/41 = 2.02. <br/> Molecular formula `= [C_(3)H_(5)]_(2) = C_(6)H_(10)` <br/> To detremine the structure of (A) and (B) : `C_(6)H_(10)` on hydrogenation with 2 moles of `H_(2)` gives `C_(6)H_(12)` and structure is 2-methylpentane. (A) on hydration in presence of dil. `H = H_(2)SO_(4)` and `HgSO_(4)` gives `C_(6)H_(12)O` which gives Iodoformtest positive. Hence, structure of `A = (CH_(3))_(2)<a href="https://interviewquestions.tuteehub.com/tag/ch-913588" style="font-weight:bold;" target="_blank" title="Click to know more about CH">CH</a>-CH-C=CH` (4-methylpentlyne) and (B) is 4-methylpentanone-2.</body></html>

Discussion

50889.	896 mL vapour of a hydrocarbon 'A' having carbon 87.80% and hydrogen 12.19% weighs 3.28 g at STP. Hydrotion of 'A' gives 2-methylpentane . Also 'A' on hydration in the presnce of H_(2)SO_(4) andHgSO_(4) gives a katone 'B' having molecular formula C_(6)H_(12)O. The ketone 'B' gives a positive iodoform test. Find the structure of 'A' and give the reactions involved.
Answer» <html><body><p></p>Solution :To <a href="https://interviewquestions.tuteehub.com/tag/determine-437910" style="font-weight:bold;" target="_blank" title="Click to know more about DETERMINE">DETERMINE</a> the molecular mass of hydrocarbon (A) 896 mL vapour of `C_(x)H_(y)` (A) weighs `3.28`g at STP <br/> 22700 mL vapour of `C_(x)H_(y)` (A) weighs `(328xx22700)/(896)g//`"mol at" STP <br/> = `83.1g//mol`<br/> Hence , molecular mass of `C_(x)H_(y)` (A) =`83.1g mol^(-)` To determine the empirical formula of hydrocarbon (A) <br/> `{:("Elament" ,%,"Atomic mass","Relative ratio","Relative no. of atoms","Somplest ratio"),(C,87.8,12,7.31,1,3),(H,12.19,1,12.19,1.66,4.98~~<a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a>):}` <br/> Thus, Empirical formula of A si `C_(3)H_(5)` . <br/> `:.` Empirical formul mass = 36 + 5 = 41. <br/> `n = ("Molecular mass")/("Emprirical formula mass")= 83.1/41 = 2.02~~ 2` <br/> Molecular mass is double of empirical formula mass. <br/> `:.` Molecular formula is `C_(6)H_(10)` <br/> To determine the structure of compounds (A) and (B) <br/><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/ARH_NCERT_EXE_CHM_XI_C13_S01_049_S01.png" width="80%"/> <br/> Hence, hydrogenation of hydrocarbon (A) requires 2 moles of hydrogen to form 2-methylpentane. Therefor, hyrocarbon (A) is an alkyne having five carbon atoms in a staight chain and a methyl substituent at position 2. Thus the possible structures for the alkyne (A) are I and II. <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/ARH_NCERT_EXE_CHM_XI_C13_S01_049_S02.png" width="80%"/> <br/> Since, addition of `H_(2)O` to alkyne (A) in presence of `Hg^(2+)` , give a ketong which gives positive iodoform test, therefore, therefore, ketone (B) must be a methyl katon , i.e., it must contain a `COOH_(3)` <a href="https://interviewquestions.tuteehub.com/tag/group-1013370" style="font-weight:bold;" target="_blank" title="Click to know more about GROUP">GROUP</a>. <br/> Now addition of `H_(2)O` to alkyne (II) should give a <a href="https://interviewquestions.tuteehub.com/tag/mixture-1098735" style="font-weight:bold;" target="_blank" title="Click to know more about MIXTURE">MIXTURE</a> of two ketones in which 2- methyl pentan -3 one (minor) and 4- methylpentan -2-one ketone (B) (which shows `+ve` iodoform test) <a href="https://interviewquestions.tuteehub.com/tag/predominates-7707028" style="font-weight:bold;" target="_blank" title="Click to know more about PREDOMINATES">PREDOMINATES</a>. <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/ARH_NCERT_EXE_CHM_XI_C13_S01_049_S03.png" width="80%"/> <br/> In contrast, addition of `H_(2)O` to alkyne (I) will give only one ketone, i.e., 4-methylpentan-2- one which gives iodoform test. <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/ARH_NCERT_EXE_CHM_XI_C13_S01_049_S04.png" width="80%"/> <br/> Thus, hydrocabon `C_(x)H_(y)` (A) is 4-methylpent -1-yne. 4-methylpentan -2 one (gives + ve iodoform test)</body></html>

Discussion

50890.	896 mL of a hydrocarbon 'A' having carbon 87.80 % and hydrogen 12.19% weighs 3.28 g at STP. Hydrogenation of 'A' gives 2-methylpentane. Also 'A' on hydration in the presence of H_(2)SO_(4) and HgSO_(4) gives a ketone 'B' having molecular formula C_(6)H_(12)O. The ketone 'B' gives a positive iodoform test. Find the structure of 'A' and give the reaction involved.
Answer» <html><body><p></p>Solution :`"896 <a href="https://interviewquestions.tuteehub.com/tag/ml-548251" style="font-weight:bold;" target="_blank" title="Click to know more about ML">ML</a> of" C_(x)H_(y) (A) "weigh"=3.28 g` <br/> `"22400 mL of" C_(x)H_(y) (A) "weigh"=(3.28)/(896) xx 22400` <br/> `=82 g mo1^(-1)` <br/> `:. "Molecular mass of " C_(x)H_(y)=82 g mo1^(-1)` <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/MOD_SPJ_CHE_XI_P2_C13_SLV_044_S01.png" width="80%"/> <br/> `"Empirical formula of A"=C_(3)H_(5)` <br/> `"Empirical formula mass"= 3 xx <a href="https://interviewquestions.tuteehub.com/tag/12-269062" style="font-weight:bold;" target="_blank" title="Click to know more about 12">12</a> +5 xx 1=41` <br/> `n=("Molecular formula mass")/("Empirical formula mass")=(82)/(41)=2` <br/> `:. "Molecular formula" =(C_(3)H_(5))_(2)=C_(6)H_(10)` <br/> Now `underset((A))(C_(6)H_(10)) overset("2 mole")to "2-Methylpentane" CH_(3)-underset(CH_(3))underset(\|)CH-CH_(2)CH_(2)-CH_(3)` <br/> The molecular has a chain of 5 carbon atoms with a <a href="https://interviewquestions.tuteehub.com/tag/methyl-1095247" style="font-weight:bold;" target="_blank" title="Click to know more about METHYL">METHYL</a> group at the second carbon atom. Since A adds a molecule of `H_(2)O` in the presence of `Hg^(2+)` and `H^(+)` to give a ketone (B), it should be an alkyne. Two possible structures of .A. are <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/MOD_SPJ_CHE_XI_P2_C13_SLV_044_S02.png" width="80%"/> <br/> Since the compounds B does not react with `AgNO_(3)` solution, the triple bond is not terminal and therefore, structure (I) is the <a href="https://interviewquestions.tuteehub.com/tag/correct-409949" style="font-weight:bold;" target="_blank" title="Click to know more about CORRECT">CORRECT</a> structure. <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/MOD_SPJ_CHE_XI_P2_C13_SLV_044_S03.png" width="80%"/></body></html>

Discussion

50891.	88 g of CO_2 are confined to a 6 L flask at 37^@C. Calculate its pressure using van der Waals' equation. Given, a = 4.17 " atm " L^2 mol^(-2), b = 0.038 L mol^(-1).
Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :8.13 <a href="https://interviewquestions.tuteehub.com/tag/atm-364409" style="font-weight:bold;" target="_blank" title="Click to know more about ATM">ATM</a></body></html>

Discussion

50892.	8.7 gm of a sample of MnO_2(molar mass = 87) is used in a chemical reaction with HCl to release Cl_2Released Cl_2is used to displace I_2from excess of KI solution. The iodine hence released is estimated using hypo and consumes 100 ml of hypo solution.12.5 ml of same hypo solution is required for complete reaction with 25 ml of 0.5 N I_2solution. Number of mili of iodine in the solution used with hypo are
Answer» <html><body><p>4.35<br/>43<br/>50<br/>6.25</p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/oxidation-588780" style="font-weight:bold;" target="_blank" title="Click to know more about OXIDATION">OXIDATION</a> <a href="https://interviewquestions.tuteehub.com/tag/number-582134" style="font-weight:bold;" target="_blank" title="Click to know more about NUMBER">NUMBER</a> of <a href="https://interviewquestions.tuteehub.com/tag/sodium-1215510" style="font-weight:bold;" target="_blank" title="Click to know more about SODIUM">SODIUM</a> in sodium <a href="https://interviewquestions.tuteehub.com/tag/amalgam-363032" style="font-weight:bold;" target="_blank" title="Click to know more about AMALGAM">AMALGAM</a> is <a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a></body></html>

Discussion

50893.	8.4 " mL of " gaseous hydrocarbon A was burnt with 50 " mL of " O_2 in a eudiometer tube. The volume of the products after cooling to room temperature was 37.4 mL. When reacted with NaOH, the volume contracted to 3.8 mL. What is the molecular formula of A.
Answer» <html><body><p></p>Solution :Let the formula of hydrocarbon `(A)=C_(x)H_(y)` <br/> `thereforeundersetunderset(8.4)(1 vol) (C_(x)H_(y)(g))+undersetunderset(8.4(x+(y)/(4))(vol)(x+(y)/(4))vol)((x+(y)/(4))O_(2)(g))toundersetunderset(8.4 x vol)(x vol)(xCO_(2)(g))+(y)/(2)H_(2)(g)` <br/> Contraction in volume `=V_(<a href="https://interviewquestions.tuteehub.com/tag/r-611811" style="font-weight:bold;" target="_blank" title="Click to know more about R">R</a>)-V_(P)` <br/> `=(8.4+50)-(37.4)` <br/> `=21mL` <br/> From equation the contraction is (volume of `H_(2)O` is not taken as it condenses) <br/> `therefore` Contraction`=(8.4[1+(cancel(x)+(y)/(4))-(cancel(x)+0)]` <br/> `implies8.4(1+(y)/(4))=21impliesy=6` <br/> After treating with `NaOH`, there is a contraction of `(37.4-3.8)=33.6mL` which is equal tot he <a href="https://interviewquestions.tuteehub.com/tag/vlume-3264243" style="font-weight:bold;" target="_blank" title="Click to know more about VLUME">VLUME</a> of `CO_(2)` produced. <br/> Volume of `CO_(2)` produced by `8.4 mL` of hydrocarbon `=8.4x` <br/> `therefore8.4x` <br/> `therefore 8.4x=33.6` <br/> `ximplies4mL` Alternate <a href="https://interviewquestions.tuteehub.com/tag/method-559432" style="font-weight:bold;" target="_blank" title="Click to know more about METHOD">METHOD</a>: <br/> After `NaOH` tretment, the volume is reduced to `3.8 mL`, which <a href="https://interviewquestions.tuteehub.com/tag/corresponds-935737" style="font-weight:bold;" target="_blank" title="Click to know more about CORRESPONDS">CORRESPONDS</a> to the volume of `O_(2)` usused. Volume of `O_(2)` (unused)`=`volume of `O_(2)` taken`-` volume of of `O_(2)` used <br/> `implies50-8.4(x+(y)/(4))=3.8` <br/> Sove to get, `y=6` <br/> Hence, hydrocarbon is `C_(4)H_(6)`.</body></html>

Discussion

50894.	83.33 gm of 117.6 % oleum is added to 1 litre of 3MK_(4)[Fe(CN)_(6)] and futher 2 litre of 33.6 volume strenght (at 1 atm, 273 K) H_(2)O is added to it reaction: K_(4)[Fe(CN)_(6)]+H_(2)SO_(4) +H_(2)O_(2) rarr K_(3)[Fe(CN)_(6)] +K_(2)SO_(4)+H_(2)O If final volume of solutin is 3 litre , then select the correct statement(s):
Answer» <html><body><p>`H_(2)SO_(4)` is <a href="https://interviewquestions.tuteehub.com/tag/limited-1074043" style="font-weight:bold;" target="_blank" title="Click to know more about LIMITED">LIMITED</a> reagent<br/>2 moles of `K_(3)Fe(<a href="https://interviewquestions.tuteehub.com/tag/cn-907325" style="font-weight:bold;" target="_blank" title="Click to know more about CN">CN</a>)_(6)` is formed<br/>If 2 ml of resulting solution is <a href="https://interviewquestions.tuteehub.com/tag/heated-1017243" style="font-weight:bold;" target="_blank" title="Click to know more about HEATED">HEATED</a> then 37.3 ml of `O_(2)` at 1 <a href="https://interviewquestions.tuteehub.com/tag/atm-364409" style="font-weight:bold;" target="_blank" title="Click to know more about ATM">ATM</a> , 273 K will be evoled from `H_(2)O_(2)` <br/>If 3 ml of resulting solution is heated then 37.3 ml of `O_(2)` at 1 atm , 273 K will be evolved from `H_(2)O_(2)` </p>Answer :A::B::C</body></html>

Discussion

50895.	8.281 has how many significant figures?
Answer» <html><body><p>1<br/>2<br/>3<br/>4</p>Answer :D</body></html>

Discussion

50896.	8.2 L of an ideal gas weights 9.0 g at 300 K and 1 atm pressure. The molecular mass of the gas is
Answer» <html><body><p>54<br/><a href="https://interviewquestions.tuteehub.com/tag/27-298706" style="font-weight:bold;" target="_blank" title="Click to know more about 27">27</a><br/>13.5<br/>81</p>Solution :27</body></html>

Discussion

50897.	8.1 g of K_2Cr_2O_7 reacts with 12.8 g of HI according to the equation Cr_2O_^(2-)+HItoCrI_3+KI+I_2 Calculate: (a). Percentage by mass of K_2Cr_2O_7 left unreasted. (b). Volume of I_2 (g) evolved, if I_2 obtained is heated to 500 K and 1.0 atm pressure.
Answer» <html><body><p></p>Solution :Balance the redox reaction: <br/> `<a href="https://interviewquestions.tuteehub.com/tag/14h-273936" style="font-weight:bold;" target="_blank" title="Click to know more about 14H">14H</a>^(o+)+cancel(6e^(-))+Cr_2O_7^(2-)to2Cr^(3+)+7H_2O` <br/> `underline(2HItoI_2+2e^(-)+2H^(+)]xx3)` <br/> `underline(8H^(o+)+Cr_2O_7^(2-)+6HIto2Cr^(3+)+3I_2+7H_2O)` <br/> To balance equation for other ions, add `2K^(o+)` and `8I^(ɵ)` to both sides. <br/> `14HI+K_2Cr_2O_7^(2-)to2CrI_3+3I_2+7H_2O` <br/> `14xx138g of <a href="https://interviewquestions.tuteehub.com/tag/hi-479908" style="font-weight:bold;" target="_blank" title="Click to know more about HI">HI</a> `requires `-=1 mol K_2Cr_2O_7-=294g` <br/> `12.8 of HI` requires`=(294xx12.8)/(14xx128)=2.1g K_2Cr_2O_7` <br/> (a). <a href="https://interviewquestions.tuteehub.com/tag/weight-1451304" style="font-weight:bold;" target="_blank" title="Click to know more about WEIGHT">WEIGHT</a> of `K_2CrO_2O_7` unreacted `=8.1-2.1=6.0g` <br/> `%` of `K_2Cr_2O_7` unreacted`=(6xx100)/(8.1)=74.07%` <br/> (<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a>). 1 " mol of "`K_2Cr_2O_7(=294g) gives =3 " mol of "I_2` <br/> `2.1 g of K_2Cr_2O_7-=(3xx2.1)/(294)=0.021 mol I_2` <br/> `PV=nRT` <br/> `V_(I_2)=(nRT)/(P)=(0.021xx0.082xx500)/(1)` <br/> `=0.861L=861 <a href="https://interviewquestions.tuteehub.com/tag/ml-548251" style="font-weight:bold;" target="_blank" title="Click to know more about ML">ML</a>`</body></html>

Discussion

50898.	80g salt of weak base and strong acid 'XY'('d' of solid XY is 2g/cc) is dissolved in water and forms 2I. Of aqueous solution. If 'XY' forms CSCl type crustal having r_(X^(@))=1.6A^(@),r_(gamma^(-))=1.864A^(@) then p^(H) of the resultant solution at 298k is [k_(b) "of" XOH=4xx10^(-5)]
Answer» <html><body><p>2<br/><a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a><br/>4<br/>5<br/></p>Solution :`d=(2M)/(a^(3)N_(0))z=1` <br/> `sqrt(3)a=100g//"mole"rArralpha=((2r_(1)+2ra))/sqrt(3)` <br/> `M=(<a href="https://interviewquestions.tuteehub.com/tag/da-943215" style="font-weight:bold;" target="_blank" title="Click to know more about DA">DA</a>^(3)N_(0))/(1)=100g//"mole"` <br/> `P^(<a href="https://interviewquestions.tuteehub.com/tag/h-1014193" style="font-weight:bold;" target="_blank" title="Click to know more about H">H</a>)=7-(1)/(2)[P^(kb)+"<a href="https://interviewquestions.tuteehub.com/tag/log-543719" style="font-weight:bold;" target="_blank" title="Click to know more about LOG">LOG</a>"c]=5`.</body></html>

Discussion

50899.	8.0575 xx 10^(-12) kg of Glauber's salt are dissolved in water to obtain 1 dm^3 of a solution of density 1077.2 kg m^(-3). Calculate the molarity.
Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :0.2502 M, 0.24 m, `4.<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a> <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> <a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>^(-3)`</body></html>

Discussion

50900.	80% carbons is present in an alkane by weight. The possible conclusions are
Answer» <html><body><p>The emperical formula of the <a href="https://interviewquestions.tuteehub.com/tag/compound-926863" style="font-weight:bold;" target="_blank" title="Click to know more about COMPOUND">COMPOUND</a> is `CH_(3)`<br/>The minimum number of carbons in the molcule is <a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a> <br/>The compound has gram atoms of C & H in 4:1 ratio <br/>This composition suits to all alkanes</p>Solution :`{:(,"C",,"H"),(,<a href="https://interviewquestions.tuteehub.com/tag/80g-338891" style="font-weight:bold;" target="_blank" title="Click to know more about 80G">80G</a>,,20g),("moles =",(80)/(<a href="https://interviewquestions.tuteehub.com/tag/12-269062" style="font-weight:bold;" target="_blank" title="Click to know more about 12">12</a>),:,(20)/(1)),(,1,:,3):}` <br/> `(n)/(2n+2)=(1)/(3)impliesn=2`</body></html>

Discussion

Explore topic-wise InterviewSolutions in Class 11.

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""(92)^(233)U is assumed to decay by emitting alpha- and beta-particles, thepossible produce of decay is

Sodium carbonate of 92% purity is used in the reation Na_(2)CO_(3)+CaCl_(2)toCaCO_(3)+2NaCl. The number of grams of Na_(2)CO_(3) required to yield 1 gm of CaCO_(3)

9.2 grams of N_(2)O_(4) (g) is taken in a closed one litre vessel and heated till the following equilibrium is reachedN_(2) O_(4) (g) hArr 2 NO_(2) (g) At equilibrium , 50% N_(2)O_(4) (g) is dissociated . What is the equilibrium constant ( in mol lit^(-1)) (" Molecular weughtof" N_(2)O_(4) = 92)

9.2 g of N_(2) O_(4 (g)) is taken in 1 lit vessel and heated . At equilibrium , 50 % is dissociated . Equilibrium constant (mol/lit) [MW = 92]

9.0g m ice at 0^(@)C is mixed with 36gm of water at 50^(@)C in a thermally insulated container, using the following data, answer the question that follow: C_(P (H_(2)O)) = 4.18 Jg^(-1) K^(-1), Delta H_("fusion") (ice) = 335 J g^(-1) Delta S_("ice") is

90g Na and 170g K is present in a person with 70 kg of body weight.

9.0g m ice at 0^(@)C is mixed with 36gm of water at 50^(@)C in a thermally insulated container, using the following data, answer the question that follow: C_(P (H_(2)O)) = 4.18 Jg^(-1) K^(-1), Delta H_("fusion") (ice) = 335 J g^(-1) Final tempeature of water is

9.0g m ice at 0^(@)C is mixed with 36gm of water at 50^(@)C in a thermally insulated container, using the following data, answer the question that follow: C_(P (H_(2)O)) = 4.18 Jg^(-1) K^(-1), Delta H_("fusion") (ice) = 335 J g^(-1) Delta S_("water") is

9.00 litres of a gas at 16 atm and 27°C weigh 93.6 g. What is the molecular mass of the gas ?

90 g of acetic acid react with excess of NaHCO_(3) then what volume of CO_(2) will produce at S.T.P. Write your answer in terms of nearest integer.

88 g of CO_2 are confined to a 6 L flask at 37^@C. Calculate its pressure using van der Waals' equation. Given, a = 4.17 " atm " L^2 mol^(-2), b = 0.038 L mol^(-1).

8.4 " mL of " gaseous hydrocarbon A was burnt with 50 " mL of " O_2 in a eudiometer tube. The volume of the products after cooling to room temperature was 37.4 mL. When reacted with NaOH, the volume contracted to 3.8 mL. What is the molecular formula of A.

8.281 has how many significant figures?

8.2 L of an ideal gas weights 9.0 g at 300 K and 1 atm pressure. The molecular mass of the gas is

8.1 g of K_2Cr_2O_7 reacts with 12.8 g of HI according to the equation Cr_2O_^(2-)+HItoCrI_3+KI+I_2 Calculate: (a). Percentage by mass of K_2Cr_2O_7 left unreasted. (b). Volume of I_2 (g) evolved, if I_2 obtained is heated to 500 K and 1.0 atm pressure.

80g salt of weak base and strong acid 'XY'('d' of solid XY is 2g/cc) is dissolved in water and forms 2I. Of aqueous solution. If 'XY' forms CSCl type crustal having r_(X^(@))=1.6A^(@),r_(gamma^(-))=1.864A^(@) then p^(H) of the resultant solution at 298k is [k_(b) "of" XOH=4xx10^(-5)]

8.0575 xx 10^(-12) kg of Glauber's salt are dissolved in water to obtain 1 dm^3 of a solution of density 1077.2 kg m^(-3). Calculate the molarity.

80% carbons is present in an alkane by weight. The possible conclusions are