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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 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) |
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Answer» `1.6xx10^(8)`M `C_(6)H_(12)O_(6)hArr 6HCHO` backward reaction `K_(2) = [(1)/(K_(1))]^(1//6)` , `K_(2) = [(1)/(6xx10^(22))]^(1//6)` `K_(2) = 1.6xx10^(-4)` M |
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| 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 : |
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Answer» 0.54 `40xx0.25=(W)/(56//2)xx1000` W=0.28 g (MASS of CaO) `% CaO=(0.28)/(0.518)xx100=54%` |
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| 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) |
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Answer» 0.040, 0.020 `= (0.50)/(98xx0.5) = 0.020` `N= Mxx2 = 0.020xx2= 0.040` |
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| 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: |
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Answer» ` 54.0% ` = no of milli equivalents of `CaC_(2)O_4` = noof milli equivalents of`KMnO_4` `= 40xx0.25 =10` No of milli moles of lime = 5 `:.` weight of `CaO = 5 xx 56 xx 10^(-3) = 0.28g` percentage of `CaO= (0.28)/(0.518)xx100=54.05%` |
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| 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. |
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Answer» `-=30.0 mL 0.1 N I_(2)` `-=30.0 mL 0.1 N Cl_(2)` `-=30.0 mL 0.1 N MnO_(2)` Amount of `MnO_(2)` present `=(NxxExxV)/(1000)` `=(1)/(10)xx(87)/(2)xx(30)/(1000)` `% MnO_(2)=(87xx30xx100)/(10xx2xx1000xx0.5)=26.1` |
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| 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. |
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Answer» Solution :`T^(@)F" of water "=273K` `T_(f)" of the solution "=272.76K` `DeltaT_(f)=T_(f^(@))-T_(f)=+0.24K` `M_(2)=(K_(f)W_(2))/(DeltaT_(f).W_(1))` OBSERVED molecular mass `M_(2)=(1.86xx0.5xx1000)/(100xx0.24)="38.75 g.mol"^(-1)` The colligative property is inversely related to the molar mass. `THEREFORE"Van.t Hoff factor"` `i=("Observed colligative property")/("Normal colligative property")` `=("Theoretical molar mass")/("Observed molar mass")` `"Vant Hoff factor (i) "=(74.5)/(38.75)=1.92` `"Degree of dissociation "alpha=(i-1)/(n-1)` `n="2 for KCL"` `therefore alpha=(1.92-1)/(2-1)` `therefore"Degree of dissociation "=0.92` |
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| 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. |
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Answer» Solution :`T^(@)F" of water "=273K` `T_(f)" of the solution "=272.76K` `DeltaT_(f)=T_(f^(@))-T_(f)=+0.24K` `M_(2)=(K_(f)W_(2))/(DeltaT_(f).W_(1))` Observed molecular mass `M_(2)=(1.86xx0.5xx1000)/(100xx0.24)="38.75 g.mol"^(-1)` The colligative property is INVERSELY related to the molar mass. `therefore"Van.t Hoff factor"` `i=("Observed colligative property")/("NORMAL colligative property")` `=("Theoretical molar mass")/("Observed molar mass")` `"Vant Hoff factor (i) "=(74.5)/(38.75)=1.92` `"Degree of dissociation "alpha=(i-1)/(n-1)` `n="2 for KCL"` `therefore alpha=(1.92-1)/(2-1)` `therefore"Degree of dissociation "=0.92` |
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| 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) |
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Answer» SOLUTION :Equivalent WEIGHT of `CuFeS_2=` molecular weight `CuFeS_2toCuSdarr+FeS` `FeS+H_2SO_4toFeSO_4+H_2Suarr` `2FeSO_4+H_2SO_4+OtoFe(SO_4)_3+H_2O` Molecular weight of `CuFeS_2=63.5+55.8+64=183.3g` `42 " ML of " 0.01 M K_2Cr_2O_7=42xx0.01xx6NK_2Cr_2O_7` `=2.52 m" Eq of "K_2Cr_2O_7` `=2.52m" Eq of "FE^(2+)` `=2.52 m" Eq of "CuFeS_2` `=2.52 m" Eq of "CuFeS_2` `=2.52xx10^(-3)xx18.3g of CuFeS_2` `=0.4619 g of CuFeS_2` `%` of `CuFeS_2=(0.4619xx100)/(0.5)=92.38%` |
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| 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 |
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Answer» Solution :Mass of the METAL = `W_(1)` = 0.456 g. Mass of the metal chloride = `W_(2)` = 0.606 g `:.` Mass of chlorine = `W_(2) -W_(1)` = 0.606 - 0.456 = 0.15 g 0.15 g of chlorine combine with 0.456 g of metal. `:.` 35.46 g of chlorine will combine with`0.456/0.15 xx 35.46` = 107.79 g `EQ ^(-1)`. `Cr_(2)O_(7)^(2-)+14H^(+)+6e^(-) rarr 2Cr^(3+)+7H_(2)O` Equivalent mass = `("Molar mass")/("No. of electrons gained")` `:.` Molar mass of `K_(2)Cr_(2)O_(7)` = 294.18 `:.` Equivalent mass of `K_(2)Cr_(2)O_(7) = 294.18/6 = 49.03` |
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| 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? |
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Answer» Solution :C=0.25 M T=370.28 K `(PI)_("gulcose")`=CRT `(pi)=0.25 mol L^(-1)xx0.082 L atm^(-1)mol^(-1) xx370.28K` |
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| 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? |
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Answer» CHLORINE |
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| 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)) |
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Answer» 6.82 mL |
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| 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. |
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Answer» |
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| 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 |
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Answer» `43.5%` = No. of m.eqts of `KI =` No. of m.eqts hypo No. of m.eqts `= MnO_(2) = 16 xx 1.25 = 16 xx (5)/(4) =20` No. of milli MOLES of `MnO_(2) = 10` WEIGHT of `MnO_(2) = 10 xx 87 xx 10^(-3) = 0.87g` weight % of `MnO_(2) = (0.87)/(1) xx 100 = 87%` |
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| 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'? |
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Answer» ` K_a = CALPHA ^(2)= 0.1 xx ( 0.01 )^(2)= 10 ^(5) ` ` K_a = ([H^(+) ] [A^(-) ]) /( [HA]) =([H^(+) ]xx 0.2 )/(0.1 )` ` RARR [H^(+) ] = calpha =10 ^(-5)xx ( 0.1)/( 0.2 ) rArr alpha = 5 xx 10 ^(-5)` |
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| 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. |
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Answer» `:. ` Concs. At equilibrium : `[CH_(3)CO OH]=0.05-0.00095 M, [CH_(3)CO O^(-)] = [H^(+)] = 0.00095 M` |
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| 50867. |
A 0.02 M solution of pyridinium hydrochloride has pH=3.44 . Calculate the ionization constantof pyridine. |
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Answer» SOLUTION :Pyridinium hydrochloride `C_6H_5N^(+) HCL^(-)` is conjugate ACID of pyridine `(C_6H_5N)`. pH=3.44 = -log `[H^+]` `therefore` log `[H^+] = -3.44 = bar4.56` `therefore [H^+] = 10^(-3.44) = 3.63xx10^(-4) = alpha` `{:(,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)) :}` `K_a=([C_6H_5NCl^(-)][H^+])/([C_6H_5N^(+)HCl^-])` `=((3.63xx10^(-4))(3.63xx10^(-4)))/0.02=6.588xx10^(-6)` Pyridine is conjugate base of pyridinium hydrochloride , so its ionization constant = `K_b` `K_b=K_w/K_a=(1.0xx10^(-14))/(6.588xx10^(-6))=1.5179xx10^(-7)` |
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| 50868. |
A 0.02 M solution of pyridinium hydrochloride has pH=3.44. Calculate the ionization constant of pyridine. |
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Answer» Solution :`pH = 3.44 , i.e., log [H^(+)]=-3.44 = bar(4).56 :.[H^(+)]=3.63xx10^(_4)M` `C_(5)H_(5)N.HCl+aq hArr C_(5)H_(5)overset(+)NCL^(-)+H^(+)` `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)` `pK_(a)=-log(6.588xx10^(-6))=6-0.8187=5.18` `pK_(a)+pK_(b)=14 :. pK_(b)=14-5.18 = 8.82` `-log K_(b) = 8.82 or log K_(b) = - 8.82 = bar(-).18 :. K_(b)=1.514xx10^(-9)` Alternatively, Pyridine hydrochloride is a salt of weak BASE and strong acid . HENCE, `pH=-(1)/(2)[log K_(w)-logK_(b)+logc],i.e., 3.44 = - (1)/(2) [-14-logK_(b)+log(2xx10^(-2))]` or `6.88=14+log K_(b) + 1.70 or log K_(b) = - 8.82 = bar(9).18 or K_(b)=1.5xx10^(-9)`. |
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| 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 |
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Answer» `-0.0093^(@)C` `KNO_(3) HARR K^(+) + NO_(3)^(-)` `therefore` Concentration in solution `=2xx0.01=0.02` `DeltaT_(f)=T_(f)-T^(@)=K_(f)xxm`for glucose `therefore k_(f)(DELTATF)/(m)=(0.018)/(0.01)=1.86^(@) m^(-1)` `DeltaT_(f)` for `KNO_(3)=1.86xx0.02=0.0372^(@)C` `T_(f)=T^(@)-T_(f)=0.0000-0.0372=-0.0372^(@)C` |
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| 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 ? |
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Answer» `K=(C alpha.C alpha)/(C (1-alpha))=(C alpha^(2))/(1-alpha)=(0.01xx(0.0134)^(2))/(1-00.01)=1.8xx10^(-6)` |
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| 50871. |
A 100 watt bulb emits monochromatic light of wavelength 400 nm . Calculate the number of photons emitted per second by the bulb. |
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Answer» |
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| 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: |
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Answer» `1:1` `:.` MOLECULAR MASS of amide `=(9.9)/(1.7) xx 17x = 99x` % NITROGEN in the amide `= (14x)/(99x) xx 100 = 14.14` Ratio of number of atoms of 'N' and 'O' `= (14.14)/(14): (33.33)/(16) = 1:2` |
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| 50873. |
99.95% dihydrogen is obtained by electrolysing warm aqueous Ba(OH)_2 solution by electrodes…….. |
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Answer» Cu |
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| 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? |
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Answer» Solution :NUMBER of milli EQUIVALENT of BASE `=200xx1=200meq`. Number of milli equivalent of acid =200m. EQ 200 milli equicalent of acid =? Equivalent weight of acid `=(9.8xx1000)/(200)=49` Basicity of acid `=("Molecular weight ")/("Equivalent weight")=(98)/(49)=2` |
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| 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 ? |
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Answer» or 35.4 (after ROUNDING off) |
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| 50876. |
""(92)^(233)U is assumed to decay by emitting alpha- and beta-particles, thepossible produce of decay is |
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Answer» `""_(82)^(207)Pb` `implies ""_(92)U^(238)` is the starting element of 4n+2 series `implies ""_(82)Pb^(206)` is the POSSIBLE DECAY PRODUCT of this series. |
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| 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) |
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Answer» 8.5g |
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| 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) |
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Answer» `0.1` At. Eqm. ( after 50% dissociation ), `[N_(2)O_(4)] = 0.05 M, [NO_(2)] = 0.1 M` ` K = ([NO_(2)])/([N_(2)O_(4)])=(0.1)^(2)/0.05 = 0.2` |
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| 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] |
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Answer» 0.1 |
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| 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 |
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Answer» `11.04 JK^(-1)` `= (9 xx 335)/(273) + 9 xx 4.18 "LN" (296.97)/(273) = 14.206` |
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| 50881. |
90g Na and 170g K is present in a person with 70 kg of body weight. |
| Answer» SOLUTION :TRUE STATEMENT | |
| 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 |
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Answer» 304.43K `=9 xx 335 + 9 xx (T - 273) xx 4.18` `48605 - 150.48T = 3015 + 37.62T - 10270.26` On SOLVING, T = 296.97K |
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| 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 |
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Answer» `-12.64 JK^(-1)` `=36 xx 4.18 "ln " (296.97)/(323) = - 12.643 JK^(-1)` |
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| 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] |
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Answer» |
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| 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 ? |
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Answer» Solution :In the present CASE, `P=16 atm, V = 9.00 L, T = 27^(@) C = 300 K, w = 93.6 g` According to the gas EQUATION, `PV = w/M RT` or `M = (wRT)/(PV)` SUBSTITUTING the values, we get `M = (93.6 xx 0.0821 xx 300)/(16 xx 9.00) = 16.01` |
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| 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. |
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Answer» |
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| 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. |
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Answer» Solution :STEP 1 . To determine the molecular mass of hydrocarbon (A). 896 mL vapours of hydrocarbon (A) weigh at STP =3.28 g `therefore` 22700 mL vapours of A will weigh at STP `=(3.28xx22700)/896 "g mol"^(-1)=83.1 g` `therefore` Molecular mass of hydrocarbon (A) = `83.1 g mol^(-1)` Step 2. To determine the empirical formula of hydrocarbon (A). `{:("ELEMENT","% age","Atomic mass","RELATIVE ratio","Relative no. of atoms","Simplest ratio"),(C,87.8,12,7.31,1,3),(H,12.19,1,12.19,1.66,5):}` Thus, empirical formula of hydrocarbon (A) =`C_3H_5` and empirical formula mass = 12 x 3 + 5 x 1 = 41 u `therefore n="Molecular mass "/"Empirical formula mass "=83.1/41=2.02 approx 2` Thus, molecular formula of hydrocarbon (A)= 2 x Empirical formula = `2 xx C_3H_5 = C_6H_10` Step 3.To determine the structures of compounds (A) and (B)  (i)Since hydrogenation of hydrocarbon (A) requires 2 MOLES 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 POSITION 2. Thus, the two possible structures for the alkyne (A) are I and II :  (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.  In contrast, addition of `H_2O` to alkyne (I) will give only one ketone , i.e., 4-methylpentan-2-one which gives iodoform test.  Thus, hydrocarbon (A) is 4-methylpent-1-yne. |
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| 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. |
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Answer» SOLUTION :To calculate molar MASS of hydrocarbon .A. it is given that 896 ml of Hydrocarbon .A. `C_(X)H_(y)` Weighs 3.28 gat STP. 22400 ml of has `AC_(x)H_(y)` mass `= 3.28 xx 22400` ml/896ml = 831 g/MOL. Hence, the molar mas of A = 831 g mol.  Thus, empirical formula `=C_(3)H_(5)` Empirical formula weight `= 3 xx 12= 36 + 5 = 41`. Molecular formula = (Empirical formula)n n = Mol. mass/Empirical mass = 831/41 = 2.02. Molecular formula `= [C_(3)H_(5)]_(2) = C_(6)H_(10)` 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)CH-CH-C=CH` (4-methylpentlyne) and (B) is 4-methylpentanone-2. |
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| 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. |
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Answer» Solution :To DETERMINE the molecular mass of hydrocarbon (A) 896 mL vapour of `C_(x)H_(y)` (A) weighs `3.28`g at STP 22700 mL vapour of `C_(x)H_(y)` (A) weighs `(328xx22700)/(896)g//`"mol at" STP = `83.1g//mol` Hence , molecular mass of `C_(x)H_(y)` (A) =`83.1g mol^(-)` To determine the empirical formula of hydrocarbon (A) `{:("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~~5):}` Thus, Empirical formula of A si `C_(3)H_(5)` . `:.` Empirical formul mass = 36 + 5 = 41. `n = ("Molecular mass")/("Emprirical formula mass")= 83.1/41 = 2.02~~ 2` Molecular mass is double of empirical formula mass. `:.` Molecular formula is `C_(6)H_(10)` To determine the structure of compounds (A) and (B)  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.  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)` GROUP. Now addition of `H_(2)O` to alkyne (II) should give a MIXTURE of two ketones in which 2- methyl pentan -3 one (minor) and 4- methylpentan -2-one ketone (B) (which shows `+ve` iodoform test) PREDOMINATES.  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.  Thus, hydrocabon `C_(x)H_(y)` (A) is 4-methylpent -1-yne. 4-methylpentan -2 one (gives + ve iodoform test) |
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| 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. |
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Answer» Solution :`"896 ML of" C_(x)H_(y) (A) "weigh"=3.28 g` `"22400 mL of" C_(x)H_(y) (A) "weigh"=(3.28)/(896) xx 22400` `=82 g mo1^(-1)` `:. "Molecular mass of " C_(x)H_(y)=82 g mo1^(-1)`  `"Empirical formula of A"=C_(3)H_(5)` `"Empirical formula mass"= 3 xx 12 +5 xx 1=41` `n=("Molecular formula mass")/("Empirical formula mass")=(82)/(41)=2` `:. "Molecular formula" =(C_(3)H_(5))_(2)=C_(6)H_(10)` 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)` The molecular has a chain of 5 carbon atoms with a METHYL 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  Since the compounds B does not react with `AgNO_(3)` solution, the triple bond is not terminal and therefore, structure (I) is the CORRECT structure. 
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| 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» SOLUTION :8.13 ATM | |
| 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 |
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Answer» 4.35 |
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| 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. |
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Answer» Solution :Let the formula of hydrocarbon `(A)=C_(x)H_(y)` `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)` Contraction in volume `=V_(R)-V_(P)` `=(8.4+50)-(37.4)` `=21mL` From equation the contraction is (volume of `H_(2)O` is not taken as it condenses) `therefore` Contraction`=(8.4[1+(cancel(x)+(y)/(4))-(cancel(x)+0)]` `implies8.4(1+(y)/(4))=21impliesy=6` After treating with `NaOH`, there is a contraction of `(37.4-3.8)=33.6mL` which is equal tot he VLUME of `CO_(2)` produced. Volume of `CO_(2)` produced by `8.4 mL` of hydrocarbon `=8.4x` `therefore8.4x` `therefore 8.4x=33.6` `ximplies4mL` Alternate METHOD: After `NaOH` tretment, the volume is reduced to `3.8 mL`, which CORRESPONDS to the volume of `O_(2)` usused. Volume of `O_(2)` (unused)`=`volume of `O_(2)` taken`-` volume of of `O_(2)` used `implies50-8.4(x+(y)/(4))=3.8` Sove to get, `y=6` Hence, hydrocarbon is `C_(4)H_(6)`. |
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| 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): |
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Answer» `H_(2)SO_(4)` is LIMITED reagent |
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| 50895. |
8.281 has how many significant figures? |
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Answer» 1 |
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| 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 |
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Answer» 54 |
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| 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. |
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Answer» Solution :Balance the redox reaction: `14H^(o+)+cancel(6e^(-))+Cr_2O_7^(2-)to2Cr^(3+)+7H_2O` `underline(2HItoI_2+2e^(-)+2H^(+)]xx3)` `underline(8H^(o+)+Cr_2O_7^(2-)+6HIto2Cr^(3+)+3I_2+7H_2O)` To balance equation for other ions, add `2K^(o+)` and `8I^(ɵ)` to both sides. `14HI+K_2Cr_2O_7^(2-)to2CrI_3+3I_2+7H_2O` `14xx138g of HI `requires `-=1 mol K_2Cr_2O_7-=294g` `12.8 of HI` requires`=(294xx12.8)/(14xx128)=2.1g K_2Cr_2O_7` (a). WEIGHT of `K_2CrO_2O_7` unreacted `=8.1-2.1=6.0g` `%` of `K_2Cr_2O_7` unreacted`=(6xx100)/(8.1)=74.07%` (B). 1 " mol of "`K_2Cr_2O_7(=294g) gives =3 " mol of "I_2` `2.1 g of K_2Cr_2O_7-=(3xx2.1)/(294)=0.021 mol I_2` `PV=nRT` `V_(I_2)=(nRT)/(P)=(0.021xx0.082xx500)/(1)` `=0.861L=861 ML` |
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| 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)] |
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Answer» 2 `sqrt(3)a=100g//"mole"rArralpha=((2r_(1)+2ra))/sqrt(3)` `M=(DA^(3)N_(0))/(1)=100g//"mole"` `P^(H)=7-(1)/(2)[P^(kb)+"LOG"c]=5`. |
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| 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. |
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Answer» |
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| 50900. |
80% carbons is present in an alkane by weight. The possible conclusions are |
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Answer» The emperical formula of the COMPOUND is `CH_(3)` `(n)/(2n+2)=(1)/(3)impliesn=2` |
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