Explore topic-wise InterviewSolutions in Class 11.

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49351.

A solution of 1 litre has 0.6g of non-radioactive Fe^(3+) with mass no. 56. To this solution 0.209g of radioactoveFe^(2+) is added with mass no. 57 and the following reaction occurred. .^(57)Fe^(2+) + .^(56)Fe^(3+) rarr .^(57)Fe^(3+)+ .^(56)Fe^(2+) At the end of one hour it was found that 10^(-5) moles of non-radioactive .^(56)Fe^(2+) mol L^(-1) hr^(-1). Negalecting any charge in volume, calculate the activity of the sample at the end of 1 hr (t_(1//2) for .^(57)Fe^(2+) = 4.62 hr.)

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> :`{:(,.^(57)Fe^(2+)+,.^(56)Fe^(3+)to,.^(57)Fe^(3+)+,.^(56)Fe^(2+)),("Before reaction",0.209/57,0.6/56,0,0),("After reaction",[0.209/57-<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/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a>)],[0.6/56-10^(-5)],10^(-5),10^(-5)):}` <br/>`.^(57)Fe^(2+)` left after reaction `= 3.667xx10^(-3) - 10^(-5)` <br/> `= 366.6xx10^(5)` mole <br/> or `N_(0) = 366.6xx10^(-5)xx6.023xx10^(23) = 2.208xx10^(<a href="https://interviewquestions.tuteehub.com/tag/21-293276" style="font-weight:bold;" target="_blank" title="Click to know more about 21">21</a>)`<br/> Also `t = (2.303)/(lambda) log (N_(0))/(N)` <br/> `1 = (2.303xx4.62)/(0.693) log (2.208xx10^(21))/(N)` <br/> `N = 1.9xx10^(21)` <br/> `:.` Rate of decay `= lambda xx N = (0.693)/(4.62) xx 1.9xx10^(21)` <br/> `= 2.85xx10^(20)` <a href="https://interviewquestions.tuteehub.com/tag/dph-2044953" style="font-weight:bold;" target="_blank" title="Click to know more about DPH">DPH</a></body></html>
Discussion
49352.

A solution of 0.1M KMnO_4is used for the reaction S_(2)O_(3)^(-2) + 2 MnO_(4)^(-) + H_(2)OrarrMnO_(2) +SO_(4)^(2-)+ OH^(-) . The volume of KMnO_4required to react 0.158gm of Na,s,o, is (MW = 158)

Answer» <html><body><p>13.33 <a href="https://interviewquestions.tuteehub.com/tag/ml-548251" style="font-weight:bold;" target="_blank" title="Click to know more about ML">ML</a> <br/>6.66 ml <br/>3.33 ml<br/> 26.67 ml </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> equivalents of `KMnO_4` = No of milli equivalents of <a href="https://interviewquestions.tuteehub.com/tag/thiosulphate-1414356" style="font-weight:bold;" target="_blank" title="Click to know more about THIOSULPHATE">THIOSULPHATE</a> ions <br/> `<a href="https://interviewquestions.tuteehub.com/tag/v-722631" style="font-weight:bold;" target="_blank" title="Click to know more about V">V</a> xx0.1 xx3=(0.158)/(((158)/<a href="https://interviewquestions.tuteehub.com/tag/8-336412" style="font-weight:bold;" target="_blank" title="Click to know more about 8">8</a>))xx1000` <br/> `V = 80/3 =26.67 ml`</body></html>
Discussion
49353.

A solution of 0.01 MCd^(2+)contains 0.01M NH_4OH. What conc. Of NH_4^(+)from NH_4Cl is necessary to prevent precipitation of Cd( OH)_2?K_(sp) " of "(OH)_2 =2.0 xx 10^(-14) ,K_b " of "NH_4 OH = 1.8 xx 10^(-5)if answer is 1.272 xx 10^(-x)mol/litre then x=________?

Answer» <html><body><p><br/></p>Solution :` K_(<a href="https://interviewquestions.tuteehub.com/tag/sp-1219706" style="font-weight:bold;" target="_blank" title="Click to know more about SP">SP</a>)=[<a href="https://interviewquestions.tuteehub.com/tag/cd-407381" style="font-weight:bold;" target="_blank" title="Click to know more about CD">CD</a>^(+2)][OH^(-) ] ^(2), 2 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 10 ^(-14)=10 ^(-2)[OH^(-)]^(2) ` <br/> ` [OH^(-) ] =<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>.414 xx 10 ^(-<a href="https://interviewquestions.tuteehub.com/tag/6-327005" style="font-weight:bold;" target="_blank" title="Click to know more about 6">6</a>)rArr K_b =([NH_4^(+) ][OH^(-) ])/( [NH_4OH]) ` <br/> ` 1.8 xx 10 ^(-5)=([ NH_4^(+)][1.4 xx 10 ^(-6)])/( 10^(-2))` <br/> ` [NH_4^(+) ]=1.2 xx 10 ^(-1) `</body></html>
Discussion
49354.

A solution of (+) 2-chloro-2-phenyl ethane in toluene racemises slowly in the presence of small amount of SbCl_(5), due to the formation of………

Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/carbanion-909137" style="font-weight:bold;" target="_blank" title="Click to know more about CARBANION">CARBANION</a> <br/>carbine<br/>for radical<br/>carbocation</p>Answer :D</body></html>
Discussion
49355.

A solution is saturated with respect to SrCO_3 and SrF_2. The [CO_3^(2-)] was found to be1.2 xx 10^(-3)M.The concentration of F^(-)in the solution would be :(K_(sp)of SrCO_3 = 7xx 10^(-10)K_(sp)of SrF_2 =8xx 10^(-10))

Answer» <html><body><p>` 1.3 <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) M` <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>.6 xx 10^(-2)M` <br/>` 3.7 xx 10^(-2) M` <br/>` 5.8 xx 10^(-7) M` </p>Solution :`[Sr^(+2) ][CO_3^(2-)] =KsP_1 ,[Sr^(+2) ][<a href="https://interviewquestions.tuteehub.com/tag/f-455800" style="font-weight:bold;" target="_blank" title="Click to know more about F">F</a>^(-) ] ^(2) = Ksp_2` <br/> `rArr [F^(-) ]^(2) =(Ksp_2)/(Ksp_1) xx [CO_3^(2-)]` <br/> ` [F^(-)]^(2)=(8xx 10^(-10))/( 7 xx 10 ^(-10)) xx 1.2 xx 10 ^(-3)` <br/> ` [F^(-)] =3.7 xx 10 ^(-2)`</body></html>
Discussion
49356.

A solution is saturated in SrCO_(3), and SrF_(2). The CO_(3)^(2-) was found to be 10^(-3) mol/L. If the concentration of Fin solution is represented as yxx 10^(-2) M then what is the value of 'y'? [Given : K_(sp) (SrCO_(3))= 2.5 xx10^(-10), K_(sp) (SrF_(2))= 10^(-10)]

Answer» <html><body><p>` 1.<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a> xx 10^(-3) M` <br/>` 2.6 xx 10^(-2)` <br/>` 3.7 xx 10^(-2) M` <br/>` 5.8 xx 10^(-7) M` </p>Solution :`[Sr^(+2) ][CO_3^(2-)] =KsP_1 ,[Sr^(+2) ][<a href="https://interviewquestions.tuteehub.com/tag/f-455800" style="font-weight:bold;" target="_blank" title="Click to know more about F">F</a>^(-) ] ^(2) = Ksp_2` <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> [F^(-) ]^(2) =(Ksp_2)/(Ksp_1) xx [CO_3^(2-)]` <br/> ` [F^(-)]^(2)=(<a href="https://interviewquestions.tuteehub.com/tag/8xx-1931282" style="font-weight:bold;" target="_blank" title="Click to know more about 8XX">8XX</a> 10^(-10))/( 7 xx 10 ^(-10)) xx 1.2 xx 10 ^(-3)` <br/> ` [F^(-)] =3.7 xx 10 ^(-2)`</body></html>
Discussion
49357.

A solution is prepared by mixing 100 ml of 0.1 MNH_(4) OH and 200 ml of 0.2 M NH_(4) Cl . pK_(b) of NH_(4) OH is 4.8 . Then the pH of the solution is

Answer» <html><body><p>`9.0` <br/>`8.0` <br/>8.5979<br/>5.4021</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
49358.

A solution is prepared by dissolving 46 g of ethyl alcohol in 90 g of water. The mole fraction of ethyl alcohol in this solution is:

Answer» <html><body><p>`46/<a href="https://interviewquestions.tuteehub.com/tag/90-341351" style="font-weight:bold;" target="_blank" title="Click to know more about 90">90</a>`<br/>`90/46`<br/>`46/(90 + 46)`<br/>`1/6`</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> :No. of moles `(n_(A))` of ethyl alcohol = `46/46=1` <br/> No. of moles `(n_(B))` of <a href="https://interviewquestions.tuteehub.com/tag/water-1449333" style="font-weight:bold;" target="_blank" title="Click to know more about WATER">WATER</a> `=90/18=5` <br/> `therefore` Mole <a href="https://interviewquestions.tuteehub.com/tag/fractioni-2657546" style="font-weight:bold;" target="_blank" title="Click to know more about FRACTIONI">FRACTIONI</a> of ethyl alcohol `=n_(A)/(n_(A) + n_(B)) = 1/(1+5) = 1/6`</body></html>
Discussion
49359.

A solution is prepared by dissolving 18.25 g NaOH in 200 mL of it. Calculate the molarity of the solution.

Answer» <html><body><p><br/></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/molarity-1100268" style="font-weight:bold;" target="_blank" title="Click to know more about MOLARITY">MOLARITY</a> of solution (M) `=(("Mass of NaOH")/("<a href="https://interviewquestions.tuteehub.com/tag/molecular-562994" style="font-weight:bold;" target="_blank" title="Click to know more about MOLECULAR">MOLECULAR</a> mass of NaOH"))/("Volume of solution in mL"/(1000))=((18.25g))/(("<a href="https://interviewquestions.tuteehub.com/tag/40-314386" style="font-weight:bold;" target="_blank" title="Click to know more about 40">40</a> g mol"^(-<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>))xx("0.2 L"))` <br/> `= 2.281 "mol L"^(-1) = 2.281 M`.</body></html>
Discussion
49360.

A solution is prepared by dissolving 2.0 g of glucose and 4.0 g urea in 100 g of water at 298 K . Calculate the vapour pressure of the solution ,If the vapour pressure of pure water is23.756 torr. (Molecular mass of urea = 60 and glucose= 180g mol^(-1))

Answer» <html><body><p>&lt;<a href="https://interviewquestions.tuteehub.com/tag/p-588962" style="font-weight:bold;" target="_blank" title="Click to know more about P">P</a>&gt;</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> :(i)`x_A =( n_A)/( n_A+n_B)` <br/> ` n_A ="No of moles <a href="https://interviewquestions.tuteehub.com/tag/water-1449333" style="font-weight:bold;" target="_blank" title="Click to know more about WATER">WATER</a> "` <br/> `n_B = "No. of <a href="https://interviewquestions.tuteehub.com/tag/molesof-2840420" style="font-weight:bold;" target="_blank" title="Click to know more about MOLESOF">MOLESOF</a> glucose " ` <br/>`n_A =( 100)/( <a href="https://interviewquestions.tuteehub.com/tag/18-278913" style="font-weight:bold;" target="_blank" title="Click to know more about 18">18</a> ) = 5.555` <br/> ` n_B =(2)/( 180) = 0.0111` <br/> Total moles ` (n_A+n_B) = 5.566` <br/> (ii)Urea <br/> ` n_A = "No. of moles of water " ` <br/> ` n_B=" No. of mole of urea " ` <br/> `n_A = 5.555` <br/> ` n_B=( 4)/( 60) = 0.0666 " moles of urea "` <br/> Total moles ` = 0.555+ 0.066 moles ` <br/> `x_B = ( n_A)/( n_A+n_B) ` <br/> `= ( 5.555)/( 5.621) = 0.9882` <br/> `P_B = P^(@) xx x_B ` <br/> `""= 23.756xx 0.9882` <br/> ` P_B = 23.42 torr`</body></html>
Discussion
49361.

A solution is prepared by dissolving 0.63 g of oxalic acid in 100 cm^(3) of water. Find the normality of the solution.

Answer» <html><body><p></p>Solution :Molecular mass of oxalic acid, <br/> `(COOH)_(2) . 2H_(2)O = 126` <br/> <a href="https://interviewquestions.tuteehub.com/tag/basicity-394484" style="font-weight:bold;" target="_blank" title="Click to know more about BASICITY">BASICITY</a> of oxalic acid =2 <br/> `therefore` <a href="https://interviewquestions.tuteehub.com/tag/equivalent-446407" style="font-weight:bold;" target="_blank" title="Click to know more about EQUIVALENT">EQUIVALENT</a> mass of oxalic acid `=126/2 = 63` <br/> Number of gram equivalents of oxalic acid <a href="https://interviewquestions.tuteehub.com/tag/dissolved-956358" style="font-weight:bold;" target="_blank" title="Click to know more about DISSOLVED">DISSOLVED</a> `=0.63/63 = 0.01`<br/> Volume of the solution `=100 cm^(3) = 100/1000 L = 0.<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a> L` <br/> <a href="https://interviewquestions.tuteehub.com/tag/normality-1124423" style="font-weight:bold;" target="_blank" title="Click to know more about NORMALITY">NORMALITY</a> of the solution `=("No. of g eq. of oxalic acid")/("Volume in litres")` <br/> `=0.01/0.1 = 0.1 g eq. L^(-1)`</body></html>
Discussion
49362.

A solution is prepared by adding 4g of a solute A to 36 of water. Calculate the mass percent of the solute.

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/mass-1088425" style="font-weight:bold;" target="_blank" title="Click to know more about MASS">MASS</a> percent `=(4)/(40) xx 100=10%`</body></html>
Discussion
49363.

A solution is prepared by adding 5g of a solute 'X' to 45 g of solvent 'Y'. What is the mass percent of the solute 'X' ?

Answer» <html><body><p>0.1<br/>0.111<br/>0.9<br/>0.75</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> :Mass <a href="https://interviewquestions.tuteehub.com/tag/percent-1150333" style="font-weight:bold;" target="_blank" title="Click to know more about PERCENT">PERCENT</a> of <a href="https://interviewquestions.tuteehub.com/tag/x-746616" style="font-weight:bold;" target="_blank" title="Click to know more about X">X</a> `=("Mass of X")/("Mass of solution")<a href="https://interviewquestions.tuteehub.com/tag/xx100-3292680" style="font-weight:bold;" target="_blank" title="Click to know more about XX100">XX100</a>` <br/> `5/(5+45)xx100`=<a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>%</body></html>
Discussion
49364.

A solution is prepared by adding 2 g of a substance A to 18 g of water. Calculate the mass percent of the solute.

Answer» <html><body><p> <br/> <br/> </p>Solution :Masspercent of A `= ("Mass of A")/("Mass of solution") <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 100` <br/> `= (<a href="https://interviewquestions.tuteehub.com/tag/2g-300351" style="font-weight:bold;" target="_blank" title="Click to know more about 2G">2G</a>)/(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a> g A + <a href="https://interviewquestions.tuteehub.com/tag/18-278913" style="font-weight:bold;" target="_blank" title="Click to know more about 18">18</a> g "of water") xx 100` <br/> `= (2g)/(20g) xx 100 = 10%`</body></html>
Discussion
49365.

A solution is prepared by adding 2 g of a substance A to 18 g of water. Calculate the massper cent of the solute

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> :Mass <a href="https://interviewquestions.tuteehub.com/tag/per-590802" style="font-weight:bold;" target="_blank" title="Click to know more about PER">PER</a> cent of `A="Mass ofA"/"Mass of solution"xx100=(2g)/("2g of A+18g of <a href="https://interviewquestions.tuteehub.com/tag/water-1449333" style="font-weight:bold;" target="_blank" title="Click to know more about WATER">WATER</a>")xx100=(2g)/(20g)xx100=10%`</body></html>
Discussion
49366.

A solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass. Calculate the mass percentage of the resulting solution.

Answer» <html><body><p></p>Solution :300 g of `25%` solution containsd solute = <a href="https://interviewquestions.tuteehub.com/tag/75-334971" style="font-weight:bold;" target="_blank" title="Click to know more about 75">75</a> g <br/> <a href="https://interviewquestions.tuteehub.com/tag/400-315233" style="font-weight:bold;" target="_blank" title="Click to know more about 400">400</a> g of `40%` solution contains solute = 160 g <br/> <a href="https://interviewquestions.tuteehub.com/tag/total-711110" style="font-weight:bold;" target="_blank" title="Click to know more about TOTAL">TOTAL</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 soute `= 160 + 75 = 235g` <br/> Total mass of solution `= 300 + 400 = 700 g` <br/> `%` of solute in the final solution `= (235)/(700) xx 100 = 33.5%` <br/> `%` of water in the final solution `= 100 - 33.5 = 66.5%`</body></html>
Discussion
49367.

A solution is made by dissolving 49g of H_2SO_4 in 250 mL of water. The molarity of the solution prepared is

Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a> M<br/>1 M<br/>4 M<br/>5 M<br/></p>Solution :Molarity `=("<a href="https://interviewquestions.tuteehub.com/tag/wt-1462289" style="font-weight:bold;" target="_blank" title="Click to know more about WT">WT</a>. of solute")/("Mol. Wt. of solute")xx(1000)/("Volume of soln. (mL)")` <br/> =`49/98xx1000/250` <br/> =<a href="https://interviewquestions.tuteehub.com/tag/2m-300757" style="font-weight:bold;" target="_blank" title="Click to know more about 2M">2M</a></body></html>
Discussion
49368.

A solution is made by dissolving 49 g of H_(2)SO_(4) in 250 mL of water. The molarity of the solution prepared is

Answer» <html><body><p>2 M<br/>1 M<br/>4 M<br/>5 M<br/></p>Solution :Molarity (M) `= ("Weight 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>")/("Molecular weight of solute")xx(<a href="https://interviewquestions.tuteehub.com/tag/1000-265236" style="font-weight:bold;" target="_blank" title="Click to know more about 1000">1000</a>)/("Volume of solution (ml))=(49)/(98)xx(1000)/(250)=2M`</body></html>
Discussion
49369.

A solution is found to contain 0.63 g of nitric acid per 100 ml of the solution . What is the pH of the solution if the acid is completely dissociated ?

Answer» <html><body><p></p>Solution :Concentrationof `HNO_(3)` solution = 0.<a href="https://interviewquestions.tuteehub.com/tag/63-330460" style="font-weight:bold;" target="_blank" title="Click to know more about 63">63</a> g <a href="https://interviewquestions.tuteehub.com/tag/per-590802" style="font-weight:bold;" target="_blank" title="Click to know more about PER">PER</a> <a href="https://interviewquestions.tuteehub.com/tag/100-263808" style="font-weight:bold;" target="_blank" title="Click to know more about 100">100</a> <a href="https://interviewquestions.tuteehub.com/tag/ml-548251" style="font-weight:bold;" target="_blank" title="Click to know more about ML">ML</a> (Given) <br/> `=6.3 g ` per litre `= (6.3)/(63) ` moles/litre = `10^(-1)M` (`:' ` Mol. Mass of `HNO_(3) = 63`) <br/> Now, `HNO_(3)`, completely ionizes as : `HNO_(3)H_(2)O rarrH_(3)O^(+)+NO_(3)^(-)` <br/> `:. [H_(3)O^(+)]=[HNO_(3)]=10^(-1)M :. pH = - log [H_(3)O^(+)]= - log 10^(-1)=1`</body></html>
Discussion
49370.

A solution is containing 2.52g "litre"^(-1) of a reductant, 25mL of this solution required 20 mL of 0.01M KMnO_(4) in acid medium for ocidation. Find the mol. Wt of reducant. Given that each of the two atoms which undergo oxidation per molecule of reducant, suffer an increasein oxidation state by one unit.

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/126-271371" style="font-weight:bold;" target="_blank" title="Click to know more about 126">126</a>`</body></html>
Discussion
49371.

A solution is formed by mixing 0.2 M NH_(4)Cl and 0.1M NH_(3). The pH of the solution will be close to (pK_(b) of ammonia solution = 4.75).

Answer» <html><body><p><br/></p>Solution :`<a href="https://interviewquestions.tuteehub.com/tag/poh-1145204" style="font-weight:bold;" target="_blank" title="Click to know more about POH">POH</a> = pK_(<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a>) + log .(["<a href="https://interviewquestions.tuteehub.com/tag/salt-1193804" style="font-weight:bold;" target="_blank" title="Click to know more about SALT">SALT</a>"])/(["<a href="https://interviewquestions.tuteehub.com/tag/base-892693" style="font-weight:bold;" target="_blank" title="Click to know more about BASE">BASE</a>"])=4.75 + log.(0.2)/(0.1)` <br/> `=4.75+ log 2 = 4.75 + 0.3010 ~=5.0` <br/> `:. <a href="https://interviewquestions.tuteehub.com/tag/ph-1145128" style="font-weight:bold;" target="_blank" title="Click to know more about PH">PH</a> = 14 - 5.0 = 9`.</body></html>
Discussion
49372.

A solution is0.01 M Kland 0. 1 M KCl . If solidAgNO_3is added to the solution, what is the [1^(-) ] when AgClbegins to precipitate[K_(sp) (Agl) =1.5 xx 10^(-16) , K_(sp)(AgCl) =1.8 xx 10^(-10)]

Answer» <html><body><p>` 3.5 xx 10^(-7)` <br/>` 6.1 xx 10^(-8) ` <br/>` 2.2 xx 10^(-7) ` <br/>` 8.3 xx 10^(-8) ` </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> :` K_(sp)=[<a href="https://interviewquestions.tuteehub.com/tag/ag-362275" style="font-weight:bold;" target="_blank" title="Click to know more about AG">AG</a>^(+) ][Cl^(-) ] ` <br/> ` [Ag^(+) ] =(1.8 xx 10^(-10))/( 10^(-1) )= 1.8 xx 10^(-<a href="https://interviewquestions.tuteehub.com/tag/9-340408" style="font-weight:bold;" target="_blank" title="Click to know more about 9">9</a>) ` <br/> ` [I^(-) ]=(K_(sp) )/( [Ag^(+) ]_(cl^(-) ))=(1.5 xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/16-276476" style="font-weight:bold;" target="_blank" title="Click to know more about 16">16</a>))/( 1.8 xx 10^(-<a href="https://interviewquestions.tuteehub.com/tag/19-280618" style="font-weight:bold;" target="_blank" title="Click to know more about 19">19</a>)) = 8.3 xx 10 ^(-8) M`</body></html>
Discussion
49373.

A solution has been prepared by dissolved 60 g of methyl alcohol in 120 g of water. What is the mole fraction of methyl alcohol and water ?

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/moles-1100459" style="font-weight:bold;" target="_blank" title="Click to know more about MOLES">MOLES</a> of `CH_(3)OH=("Mass of "CH_(3)OH)/("<a href="https://interviewquestions.tuteehub.com/tag/molar-562965" style="font-weight:bold;" target="_blank" title="Click to know more about MOLAR">MOLAR</a> mass")=((60 <a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a>)/("<a href="https://interviewquestions.tuteehub.com/tag/32-307188" style="font-weight:bold;" target="_blank" title="Click to know more about 32">32</a> g mol"^(-1)))=1.875` mol <br/> Moles of water `= ("Mass of water")/("Molar mass")=((120g)/("18 g mol"^(-1)))=6.667` mol <br/> Mole fraction of `CH_(3)OH=(("1.875 mol"))/(("1.875 mol +6.667 mol"))=(("1.875 mol"))/(("8.542 mol"))=0.220` <br/> Mole fraction water `= 1 - 0.220 = 0.780`.</body></html>
Discussion
49374.

A solution gives yellow colour with orange, methyl red and phenol red. What is the pH of the solution ?

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> :Yellow colour with methyl <a href="https://interviewquestions.tuteehub.com/tag/orange-1138097" style="font-weight:bold;" target="_blank" title="Click to know more about ORANGE">ORANGE</a> <a href="https://interviewquestions.tuteehub.com/tag/means-1091780" style="font-weight:bold;" target="_blank" title="Click to know more about MEANS">MEANS</a> `<a href="https://interviewquestions.tuteehub.com/tag/ph-1145128" style="font-weight:bold;" target="_blank" title="Click to know more about PH">PH</a> gt 4.5` <br/> Yellow colour with methyl redmeans pH ` gt6.2` <br/> Yellow colour with phenol red means `pH lt 6.4` . Hence, thesolution has pH between 6.2 to 6.4.</body></html>
Discussion
49375.

A solution cotaning 0.10 M in Ba(NO_(3))_(2) and 0.10 M in Sr(NO_(3))_(2). If solid Na_(2)CrO_(4) is added to the solution, what is [Ba^(2+)], when SrCrO_(4) beings to precipitate? [K_(sp)(BaCrO_(4)) = 1.2 xx 10^(-10), K_(sp) (SrCrO_(4)) = 3.5 xx 10^(-5)]

Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/7-332378" style="font-weight:bold;" target="_blank" title="Click to know more about 7">7</a>.4 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 10^(-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>.0 xx 10^(-7)`<br/>`6.1 xx 10^(-7)`<br/>`3.4 xx 10^(-7)`</p>Solution :`K_(<a href="https://interviewquestions.tuteehub.com/tag/sp-1219706" style="font-weight:bold;" target="_blank" title="Click to know more about SP">SP</a>)(SrCrO_(4)) = [Sr^(2+)][CrO_(4)^(2-)]` <br/> `[CrO_(4)^(2-)] = (3.5 xx 10^(-5))/(0.1)= 3.5 xx 10^(-4)` <br/> `K_(sp)(BaCrO_(4)) = [Ba^(2)][CrO_(4)^(2-)]` <br/> `[CrO_(4)^(2-)]_("total") = [CrO_(4)^(2-)]` from `SrCrO_(4)` <br/> `[Ba^(2+)] = (1.2 xx 10^(-10))/(3.5 xx 10^(-4)) = 3.4 xx 10^(-7)`</body></html>
Discussion
49376.

A solution contians 0.1 M NaCl, 0.01 M NaBr and 0.001 M Nal.solid AgNO_(3)is gradually added to the solution and its addition K_(sp)AgCl = 10^(-10), K_(sp) AgBr = 10^(-13), K_(sp)AgI = 10^(-17) choose the correct statements

Answer» <html><body><p>`AgI` <a href="https://interviewquestions.tuteehub.com/tag/precipitates-1162809" style="font-weight:bold;" target="_blank" title="Click to know more about PRECIPITATES">PRECIPITATES</a> first<br/>The maximum `[I^(-)]` which can be maintained in the solution so that only `AgI` is precipitate is `10^(-6) M`.<br/>When the precipitation of `<a href="https://interviewquestions.tuteehub.com/tag/agcl-368919" style="font-weight:bold;" target="_blank" title="Click to know more about AGCL">AGCL</a>` just starts, `[I^(-)]` in the solution is `10^(-8) M`<br/>If sufficient NaIis added to a saturated solutions of `AgCl`, precipitate of `Agl` is obtained.</p>Solution :`{:(10ml(N)/(<a href="https://interviewquestions.tuteehub.com/tag/20-287209" style="font-weight:bold;" target="_blank" title="Click to know more about 20">20</a>)<a href="https://interviewquestions.tuteehub.com/tag/naoh-572531" style="font-weight:bold;" target="_blank" title="Click to know more about NAOH">NAOH</a> +,20ml(N)/(20)HCl),("milli eq."=10 xx 1/20 = 1/2,"milli eq." = 20 xx 1/20 = 1):}` <br/>Solution also turn s methly orange red<br/> `[H^(+)] gt [H^(-)]` <br/> then (B) (C) and (D) are <a href="https://interviewquestions.tuteehub.com/tag/correct-409949" style="font-weight:bold;" target="_blank" title="Click to know more about CORRECT">CORRECT</a>.</body></html>
Discussion
49377.

A solution contains Na_(2)CO_(3) " and " NaHCO_(3), 10 mL of this solution required 2.5 mL of 0.1 M H_(2)SO_(4) for neutralisation using phenolphthalein indicator. Methyl orange is added after first end point, further titration required 2.5 mL of 0.2 M H_(2)SO_(4). The amount of Na_(2)CO_(3) " and " NaHCO_(3) in 1 litre of the solution is :

Answer» <html><body><p>5.3 <a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a> and 4.2 g<br/>3.3 g and 6.2 g<br/>4.2 g and 5.3 g<br/>6.2 g and 3.3 g</p>Solution :N//A</body></html>
Discussion
49378.

A solution contains Na_(2)CO_(3) " and " NaHCO_(3). 10 mL of the solution required 2.5 mL of 0.1 MH_(2)SO_(4) forneutralisation using phenolphthalein as indicator. Methyl orange is then added when a further 2.5 mL of 0.2 M H_(2)SO_(4) was required. Calculate the amount of Na_(2)CO_(3) " and" NaHCO_(3) in one litre of the solution.

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/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>.5 mL of `0.1 M H_(2)SO_(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>)=2.5 mL " of" 0.2 N H_(2)SO_(4)` <br/> `=(1)/(2)Na_(2)CO_(3)` present in 10 mL of mixture <br/> So, <br/> `5 mL " of" 0.2 N H_(2)SO_(4)=Na_(2)CO_(3)` present in 10 mL of mixture <br/> `-=5 mL " of" 0.2 N Na_(2)CO_(3)` <br/> `-=(0.2xx53)/(1000)xx5=0.053 g` <br/> Amount of `Na_(2)CO_(3)=(0.053)/(10)xx1000=5.3 g//L` of mixture <br/> Between first and second end <a href="https://interviewquestions.tuteehub.com/tag/points-1157347" style="font-weight:bold;" target="_blank" title="Click to know more about POINTS">POINTS</a>, <br/> =2.5 mL of `0.2 M H_(2)SO_(4)` used <br/> =2.5 mL of `0.4N H_(2)SO_(4)` used <br/> =5 mL of `0.2 N H_(2)SO_(4)` used <br/> `-=(1)/(2)Na_(2)CO_(3)+NaHCO_(3)` present in 10 mL of mixture <br/> `(5-2.5)mL 0.2 N H_(2)SO_(4)` <br/> `-=NaHCO_(3)` present in 10 mL of mixture <br/> `-=2.5 mL 0.2 N NaHCO_(3)` <br/> `-=(0.2xx84)/(1000)xx2.5=0.042 g` <br/> Amount of `NaHCO_(3)=(0.042)(10)xx1000=4.20 g//L` of mixture.</body></html>
Discussion
49379.

A solution contains Na_(2)CO_(3) " and " NaHCO_(3). 10 mL of the solution required 2.5 mL of 0.1 M H_(2)SO_(4) for neutralisation using phenophthalein as indicator. Methyl orange is then added when a further 2.5 mL of 0.2 M H_(2)SO_(4) was required . Then the amount of Na_(2)CO_(3) " and " NaHCO_(3) in 1 litre of the solution is :

Answer» <html><body><p>5.3 <a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a> and 4.2 g<br/>3.3 g and 6.2 g<br/>4.2 g and 5.3 g<br/>6.2 g and 3.3 g</p>Solution :N//A</body></html>
Discussion
49380.

A solution contains mixture of H_(2)SO_(4), H_(2)C_(2)O_420 ml of this solution requires 40 ml of M/10NaOH for neutralization and 20 ml of N/10 KMnO_4 for oxidation. The molarity of H_2C_2O_4, H_2SO_4 are

Answer» <html><body><p>` <a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</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>, 0.1`<br/>`0.1, 0. <a href="https://interviewquestions.tuteehub.com/tag/05-256274" style="font-weight:bold;" target="_blank" title="Click to know more about 05">05</a>` <br/>` 0.05, 0.1`<br/>` 0.05, 0.05`</p>Solution :Let molarity of `H_(2)SO_(4)= M_1 and H_2C_2O_4=M_2` <br/> No. of milli equivalents of `(H_2SO_4+H_2C_2O_4)` = No. 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 NaOH <br/> `20(2M_1+2M_2)=40xx1/10` <br/> `<a href="https://interviewquestions.tuteehub.com/tag/40-314386" style="font-weight:bold;" target="_blank" title="Click to know more about 40">40</a>(M_1+M_2)=4` <br/> `M_1+M_2=0.1 ""...(1)` <br/> No.of milli equivalents of `H_(2)C_(2)O_(4)=` No. of mili equivalents `KMnO_4` <br/> `20(2M_2)=20 xx1/10` <br/> `M_(2) =1/20 =0.05 M...(2)` <br/> `M_1 =0.1 -0.05 =0.05M`</body></html>
Discussion
49381.

A solution contains equimolar concentration of a weak acid HA and its conjugate base A^(-) , p K_(b) of A^(-) is 9 . The pH of the solution is

Answer» <html><body><p>9<br/>5<br/>7<br/>`5.301`</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/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a></body></html>
Discussion
49382.

A solution contains a mixture of isotopes of X^(A)(t_(1//2) = 14 days) and X^(A_(2)) (t_(1//2) = 25 days(. Total acticity is 1 curie at t = 0. The activity reduces by 50% in 20 days. Find: (a) The initial activitites of X^(A_(1)) and X^(A_(2)) (b) The ratio of theirinitial no.of nuclei.

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) `3.3669 <a href="https://interviewquestions.tuteehub.com/tag/ci-408488" style="font-weight:bold;" target="_blank" title="Click to know more about CI">CI</a>, 0.6331 Ci`, (<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a>) `0.3245`</body></html>
Discussion
49383.

A solution contains 410.3 g of H_2SO_4 per litre of the solution at 20°C. If its density is 1.243 g cm^(-3), what will be its molality and 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> : 5.028 m, 4.187 M</body></html>
Discussion
49384.

A solution contains 25% water, 25% ethanol and 50% acetic acid by mass. Calculate the mole fraction of each component.

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> :Suppose, we have 100 g of the <a href="https://interviewquestions.tuteehub.com/tag/given-473447" style="font-weight:bold;" target="_blank" title="Click to know more about GIVEN">GIVEN</a> solution. Then, <br/> mass of <a href="https://interviewquestions.tuteehub.com/tag/water-1449333" style="font-weight:bold;" target="_blank" title="Click to know more about WATER">WATER</a> = 25g mass of ethanol = 25 g and mass of acetic acid = 50 g. <br/> The molecular <a href="https://interviewquestions.tuteehub.com/tag/masses-1088822" style="font-weight:bold;" target="_blank" title="Click to know more about MASSES">MASSES</a> of water, ethanol and acetic acid are 18, 46 and 60 <a href="https://interviewquestions.tuteehub.com/tag/respectively-1186938" style="font-weight:bold;" target="_blank" title="Click to know more about RESPECTIVELY">RESPECTIVELY</a>. Therefore,<br/> number of moles of water `(n_(1)) = 25/18 = 1.39` <br/> number of moles of ethanol `(n_(2)) = 25/46 = 0.54` <br/> and number of moles of acetic acid `(n_(3)) = 50/60 = 0.83` <br/> Total number of moles in solution = `n_(1) + n_(2) + n_(3) = 1.39 + 0.54 + 0.83 = 2.76`<br/> Mole fraction of water = `n_(1)/(n_(1) + n_(2)+ n_(3))` <br/> `=1.39/2.76 = 0.503` <br/> Mole fraction of ethanol `=n_(2)/(n_(1) + n_(2) + n_(3))` <br/> `=0.54/2.76 = 0.196` <br/> Mole fraction of acetic acid = `n_(3)/(n_(1) + n_(2) + n_(3)) = 0.83/2.76 = 0.301`</body></html>
Discussion
49385.

A solution contains 20 g of sodium chloride in 95 cm^(3) solution. The density of solution is 1.25 "g cm"^(-3). What is the mass percent of NaCl ?

Answer» <html><body><p><br/></p>Solution :Density of <a href="https://interviewquestions.tuteehub.com/tag/nacl-572483" style="font-weight:bold;" target="_blank" title="Click to know more about NACL">NACL</a> solution `= 1.25 "<a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a> cm"^(-3)` , Volume of solution `= 95 cm^(3)` <br/> <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 solution = volume `xx` density `= ("95 cm"^(3))xx("1.25 g cm"^(-3))=118.75`g <br/> Mass <a href="https://interviewquestions.tuteehub.com/tag/percent-1150333" style="font-weight:bold;" target="_blank" title="Click to know more about PERCENT">PERCENT</a> of NaCl `= ("Mass of NaCl")/("Mass of solution")xx100` <br/> `=((20g))/((118.75g))xx100=16.84%`.</body></html>
Discussion
49386.

A solutioncontains 1 milli-curiefo L- phenyol alanine C^(14)(unifromly labellled) in 2.0mL solution. The activity of labelled sample is given as 150 milli-curie/milli-mole. Calculatate: (a) The concentration of sample in the solution inmol/litre. (b) The actitivity of the solution in terms of counting per minute/mL at a countingefficiencyof 80%.

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) `3.33xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)M`, (<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a>) `88.8xx10^(<a href="https://interviewquestions.tuteehub.com/tag/7-332378" style="font-weight:bold;" target="_blank" title="Click to know more about 7">7</a>) cpm//mL`</body></html>
Discussion
49387.

A solution contains 1 g mol. Each of p-toluene diazonium chloride and p-nitrophyenl diazonium chloride. To this 1g mol. Of alkaline of phenol is added. Predict the major product. Explain you answer.

Answer» <html><body><p></p>Solution :This reaction is an example of electrophilic <a href="https://interviewquestions.tuteehub.com/tag/aromatic-363924" style="font-weight:bold;" target="_blank" title="Click to know more about AROMATIC">AROMATIC</a> substitution. In alkaline <a href="https://interviewquestions.tuteehub.com/tag/medium-1092763" style="font-weight:bold;" target="_blank" title="Click to know more about MEDIUM">MEDIUM</a>, phenol generates phenoxide <a href="https://interviewquestions.tuteehub.com/tag/ion-1051153" style="font-weight:bold;" target="_blank" title="Click to know more about ION">ION</a> which is more electron rich than phenol and hence more reactive for electrophilic <a href="https://interviewquestions.tuteehub.com/tag/attack-887517" style="font-weight:bold;" target="_blank" title="Click to know more about ATTACK">ATTACK</a>. The electrophile ini this reaction is aryladizonium cation. Stronger the electrophile faster is the reaction. p Nitrophenyldiazonium cation is a stronger electrophile than p-toluene diazonium cation. Therefore, it couples preferntially with phenol. <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/RES_CHM_AC_E01_018_S01.png" width="80%"/></body></html>
Discussion
49388.

A solution contains 0.1M H_2S and 0.3M HCI. If K_a and K_(a2) for H_2 S are 1 xx 10^(-7) and 1.3 xx 10^(-13) respectively, calculate concentration of HS^- and S^- in the mixture.

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> :First ionisation, step 1 `,H_2 ShArr H^(+) +<a href="https://interviewquestions.tuteehub.com/tag/hs-492079" style="font-weight:bold;" target="_blank" title="Click to know more about HS">HS</a>^(-) """ K_a = 1 xx 10^(-<a href="https://interviewquestions.tuteehub.com/tag/7-332378" style="font-weight:bold;" target="_blank" title="Click to know more about 7">7</a>)` <br/> Second ionisation, step 2,`HS^(-)hArr H^(+)+s^(2-) "" K_(a_2) = 3 xx 10^(-13)` <br/> Presence of HCI supresses the dissociation of `H_2 S` due to common ion effect of `H^(+)`. <br/> For step `1,1 xx 10^(-7)= ([H^+][HS^(-)])/( [H_2 S])=(0.3[HS^(-)])/( 0.1)` <br/> Cocentrationof `HS^(-) `,`[HS^(-)])= 3.3 xx 10^(-<a href="https://interviewquestions.tuteehub.com/tag/8-336412" style="font-weight:bold;" target="_blank" title="Click to know more about 8">8</a>)mol L^(-1)`<br/> or step` 2,1,.3 xx 10^(-13) = ([H^+] [S^(2-)])/([HS^(-)]) = ( 0.3 [S^(2-)])/(3.3 xx 10^(-8))` <br/> concentrationof `s^(2-) , [S^(2-)] = 1.43 xx 10^(-20) mol L^(-1)`</body></html>
Discussion
49389.

A solution contains 0.05M of each of NaCl and Na_(2),CrO_(4),. Solid AgNO_(3), is gradually added to it. Whichof the following facts true (Given: K_(sp)(AgCl) = 1.7xx 106(-10) M^(2) and K_(sp)(Ag_(2),CrO_(4),) = 1.9 xx 10^(-12) M^(3):

Answer» <html><body><p>`Cl^(-) ` ions are precipitated first <br/>`CrO_4^(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>-)` ions are precipitated first <br/>Both `Cl^(-) and CrO_4^(2-)` ions are precipitated <a href="https://interviewquestions.tuteehub.com/tag/together-3221943" style="font-weight:bold;" target="_blank" title="Click to know more about TOGETHER">TOGETHER</a> <br/>The second ion starts <a href="https://interviewquestions.tuteehub.com/tag/precipitating-2948233" style="font-weight:bold;" target="_blank" title="Click to know more about PRECIPITATING">PRECIPITATING</a> when ` [1^(st) ion ] =2.758 xx 10^(-5)` </p>Solution :` (a) [Ag^(+) ] _(cl^(-)) = ( 1.7 xx 10^(_10) )/(5xx 10 ^(_2)) =3.4 xx 10 ^(_9) M` <br/> ` [Ag^(+) ]_(CrO_4^(_2) ) = sqrt( (1.9 xx 10^(-12))/(5 xx 10 ^(_2)))=6.1 xx 10 ^(_6) M` <br/> ` Cl^(_) `Starts precipitation first<br/>` [Cl^(-)] =(K_(<a href="https://interviewquestions.tuteehub.com/tag/sp-1219706" style="font-weight:bold;" target="_blank" title="Click to know more about SP">SP</a>))/( [Ag^(+)]_(CrO_4^(_2)))= (1.7 xx 10 ^(-10))/(6.1 xx 10^(_6)) ` <br/> ` ""= 2.78 xx 10 ^(_5) M`</body></html>
Discussion
49390.

A solution containing NH_(4)Cl and NH_(4)OH has [OH]=10^(-6) mol L^(-1), which of the following hydroxides would be precipitated when this solution in added in equal volume to a solution containing 0.1 M of metal ions?

Answer» <html><body><p>`Mg(OH)_(2) (K_(<a href="https://interviewquestions.tuteehub.com/tag/sp-1219706" style="font-weight:bold;" target="_blank" title="Click to know more about SP">SP</a>)=3xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/11-267621" style="font-weight:bold;" target="_blank" title="Click to know more about 11">11</a>))`<br/>`Fe(OH)_(2) (K_(sp)=8xx10^(-16))`<br/>`Cd(OH)_(2) (K_(sp)=8xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/6-327005" style="font-weight:bold;" target="_blank" title="Click to know more about 6">6</a>))`<br/>`AgOH(K_(sp)=5xx10^(-3))`</p>Solution :When equal volumes of `(NH_(4)Cl+NH_(4)OH)` and metal ions are mixed (volume becomes double and concentration is halved) <br/> `:. [OH]=(10^(6))/(2),[M^(+<a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a>)]=(0.1)/(2)` <br/> `Q_(sp)` ( or `IP`) of metal hydroxides of `M(OH)_(2)` type <br/> `=[M^(+n)][OH]^(2)` <br/> `=((0.1)/(2))((10^(-6))/(2))^(2)=(1)/(8)xx10^(-13)` <br/> `=0.125xx10^(-13)=12.5xx10^(-11)` <br/> `:. Q_(sp) lt K_(sp)` of `Mg(OH)_(2) (1.25xx10^(-11)gt3xx10^(-11))` <br/> and `Q_(sp)gtK_(sp)` of `Fe(OH)_(2) (1.25xx10^(-11)gt8xx10^(-16))` <br/> So both can be precipitated. Since the `K_(sp)` of `Fe(OH)_(2)` is less than `K_(sp)` of `Mg(OH)_(2)`, so `Fe(OH)_(2)` will be precipitated <a href="https://interviewquestions.tuteehub.com/tag/first-461760" style="font-weight:bold;" target="_blank" title="Click to know more about FIRST">FIRST</a>. Similarly, `Q_(sp)` (or `IP`) of metal hydroxides of `M(OH)` type <br/> `=[M^(+1)][OH]=((0.1)/(2))((10^(-6))/(2))=0.25xx10^(-7)` <br/> `= Q_(sp)ltK_(sp)` of `AgOH(0.25xx10^(-7)lt5xx10^(-3))` <br/> It cannot be precipitated out. Hence `Fe(OH)_(2)` will be precipitated</body></html>
Discussion
49391.

A solution containing Na_(2)CO_(3) and NaOH requires 300 mL of 0.1 NHCl using phenolphthalein as an indicator. Methyl orange is then added to above titrated solution when a further 25 mL of 0.2 M HCl is required. The amount of NaOH present in the original solution is

Answer» <html><body><p>0.5 <a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a><br/>1g<br/>2g<br/>4g</p>Solution :300 ml of 0.1 N HCl will neutralize all the <a href="https://interviewquestions.tuteehub.com/tag/naoh-572531" style="font-weight:bold;" target="_blank" title="Click to know more about NAOH">NAOH</a> and <a href="https://interviewquestions.tuteehub.com/tag/half-1014510" style="font-weight:bold;" target="_blank" title="Click to know more about HALF">HALF</a> of `Na_(2)CO_(3)`(with phenolphthalein asindicator). The remaining half will be neutralized by 25 ml of 0.2 N HCl (with methyl orange as indicator). Thus, 0.2 N HCl required for half neutralization of `Na_(2)CO_(3) = 25 ml` or 0.1 N HCl required will be 50 ml. Hence, 0.1 N HCl <a href="https://interviewquestions.tuteehub.com/tag/used-2318798" style="font-weight:bold;" target="_blank" title="Click to know more about USED">USED</a> up for NaOH = 300 - 50 = 250 ml . Gram equivalents present in 250 ml of 0.1 N HCl `=(0.1)/(1000)xx250=0.025`. It will neutralize 0.025 g eq. of NaOH, i.e., `0.025xx40g` = 1 g.</body></html>
Discussion
49392.

A solution containing Na_(2)CO_(3) and NaOH requires 300 mL of 0.1 N HCl using phenolphthalein as an indicator. Methyl orange is then added to above titrated solution when a further 25 mL of 0.2 N HCl is required. The amount of NaOH present in the original solution is :

Answer» <html><body><p>0.5 g<br/>1 g<br/>2 g<br/>4 g</p>Solution :300 mL HCl of 0.1 <a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a> <a href="https://interviewquestions.tuteehub.com/tag/neutralises-2194775" style="font-weight:bold;" target="_blank" title="Click to know more about NEUTRALISES">NEUTRALISES</a> entire amount of NaOH and `1//2 " of" Na_(2)CO_(3)`. Remaining `1//2 " of " Na_(2)CO_(3)` is neutralised by 25 mL 0.2 N HCl, i.e., 50 mL of 0.1 N HCl. <br/> <a href="https://interviewquestions.tuteehub.com/tag/thus-2307358" style="font-weight:bold;" target="_blank" title="Click to know more about THUS">THUS</a>, 250 mL of 0.1 N HCl is required to neutralise NaOH completely. <br/> `N_(1)V_(1)(NaOH)=N_(2)V_(2)(HCl)` <br/> `=0.1xx250` <br/> =25 <br/> `W_(NaOH)=(ENV)/(1000)=(40xx25)/(1000)=1 g`</body></html>
Discussion
49393.

A solution containing Fe^(2+) ions is titrated with KMnO_(4) solution, Indicator used will be :

Answer» <html><body><p>phenolphthalein<br/>methyl orange<br/>litmus<br/>none of these</p>Solution :N//A</body></html>
Discussion
49394.

A solution containing Cu^(2+) and C_(2)O_(4)^(2-) ions is titrated with 20 " mL of " (M)/(4) KMnO_(4) solution in acidic medium. The resulting solution is treated with excess of KI after neutralisation. The evolved I_(2) is then absorbed is 25 " mL of " (M)/(10) hypo solution. Which of the following statements are correct?

Answer» <html><body><p>The difference of the number of m " mol of "`Cu^(2+)` and `C_(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>)^(2-)` ions in the <a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> is 10 m mol<br/>The difference of the number of m " mol of "`Cu^(2+)` and `C_(2)O_(4)^(2-)` ions in the solution is 22.5 m mol.<br/>The equivalent weight of `Cu^(2+)` ions in the titration with KI is equal to the atomic weight of `Cu^(2+)`<br/>The equivalent weight of KI in the titratio is `(M)/(2)` `(M=` molesular weight of KI)</p>Solution :(i). `Cu^(2+)` does not react with `MnO_(4)^(ɵ)` <br/> Only `C_(2)O_(4)^(2-)` reacts with `MnO_(4)^(2-)` <br/> `MnO_(4)^(ɵ)=C_(2)O_(4)^(2-)` <br/> `(n=5)(n=2)` <br/> `mEq-=mEq` <br/> `20xx(M)/(4)xx5-=mEq` <br/> `thereforem" Eq of "C_(2)O_(4)^(2-)=25` <br/> `mmoles of `C_(2)O_(4)^(2-)=(25)/(2)=12.5` <br/> (<a href="https://interviewquestions.tuteehub.com/tag/ii-1036832" style="font-weight:bold;" target="_blank" title="Click to know more about II">II</a>). `Cu^(2+)=KI-=I_(2)=S_(2)O_(3)^(2-)` (hypo) <br/> `mEq-=mEq-=mEq-=mEq` <br/> `(2Cu^(2+)+2I^(ɵ)toCu_(2)I_(2))(2S_(2)O_(3)^(2-)to_S_(4)O_(6)^(2-)+2e^(-))` <br/> `(e+Cu^(2+)toCu^(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>+))` <br/> `m" Eq of "S_(2)O_(3)^(2-)-=25xx(M)/(10)xx1` <br/> `=2.5m" Eq of "Cu^(2+)` <br/> `m" Eq of "Cu^(2+)=2.5` <br/> `mmoles of Cu^(2+)=(2.5)/(1)=2.5` <br/> Difference in mmoles of `C_(2)O_(4)^(2-)` and `Cu^(2+)=12.5-2.5=10` <br/> <a href="https://interviewquestions.tuteehub.com/tag/ew-447045" style="font-weight:bold;" target="_blank" title="Click to know more about EW">EW</a> of `Cu^(2+)=("atomic weight" of Cu^(2+))/(1("n-factor"=1))` <br/> Ew of `KI=(M)/("nfactor")=(M)/(1)=M{:(2I^(ɵ)toI_(2)+2e^(-)),(n=(2)/(2)=1):}`</body></html>
Discussion
49395.

A solution containing a weak acid and its conjugate acts as an acidic buffer. In a buffer the dissociation of the well acid is suppressed by its conjugate base. (K_(1),K_(2), and K_(3), of H_(3)PO_(4), are 10_(-4), 10^(-4), 10^(-13) respectively What is the P^(H) of a solution obtained by mixing 100ml of 0.1 MH(3) PO_(4), and 150ml of 0.IM NaOH

Answer» <html><body><p>`8` <br/>` <a href="https://interviewquestions.tuteehub.com/tag/9-340408" style="font-weight:bold;" target="_blank" title="Click to know more about 9">9</a>` <br/>` 7` <br/>`6` </p>Solution :` H_3 PO_4 + NaOH to NaH_2PO_4 +H_2O ` <br/>` {:( <a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a> m " <a href="https://interviewquestions.tuteehub.com/tag/moles-1100459" style="font-weight:bold;" target="_blank" title="Click to know more about MOLES">MOLES</a> " , <a href="https://interviewquestions.tuteehub.com/tag/15-274069" style="font-weight:bold;" target="_blank" title="Click to know more about 15">15</a> m " moles" ,0) , ( - ,10" moles " , 10 "moles " ), (- , 5 " moles " , 10 m " moles ") :} ` <br/> ` NaH_2PO_4 + NaOH to Na_2HPO_4 + H_2O `<br/> ` {:( 10 m " moles" , 5 m " moles ",0),( 5, - , 5 m " moles " ),( 5 m " moles" ,- , 5 m " moles") :} ` <br/> ` pH =P^(K_(<a href="https://interviewquestions.tuteehub.com/tag/a2-844567" style="font-weight:bold;" target="_blank" title="Click to know more about A2">A2</a>)) + log ""(S)/(A)rArr pH = 8`</body></html>
Discussion
49396.

A solution containing a weak acid and its conjugate acts as an acidic buffer. In a buffer the dissociation of the well acid is suppressed by its conjugate base. (K_(1),K_(2), and K_(3), of H_(3)PO_(4), are 10_(-4), 10^(-4), 10^(-13) respectively Which of the following volume of 0.1 M NaOH added to 100 mlof 0.1MH_(3),PO_(4), does not form a buffer

Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/50-322056" style="font-weight:bold;" target="_blank" title="Click to know more about 50">50</a> ml <br/>150 ml <br/>200 ml <br/>250ml </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> :In the above question if we <a href="https://interviewquestions.tuteehub.com/tag/used-2318798" style="font-weight:bold;" target="_blank" title="Click to know more about USED">USED</a> 200 ml, 0.1 MNaOH, only <a href="https://interviewquestions.tuteehub.com/tag/salt-1193804" style="font-weight:bold;" target="_blank" title="Click to know more about SALT">SALT</a>`Na_2HPO_4` is formed</body></html>
Discussion
49397.

A solution containing 8 g of a substance in 100 g of diethyl ether boils at 36.86^(@)C, whereas pure ether boils at 35.60^(@)C. Determine the molecular mass of the solute (For ether K_(b)= 2.02 K kg 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> :We have, 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>,` W_(2)=8g` <br/> Mass of solvent, `W_(1) = 100 g `<br/> Elevation of boiling point, `Delta T_(b) = 36.86 - 35.60=1.26 ^(@)C` <br/> `K_(b)=2.02` <br/> Molecular mass of the solute `M _(2) = ( <a href="https://interviewquestions.tuteehub.com/tag/1000-265236" style="font-weight:bold;" target="_blank" title="Click to know more about 1000">1000</a> xx W_(2) xx <a href="https://interviewquestions.tuteehub.com/tag/k-527196" style="font-weight:bold;" target="_blank" title="Click to know more about K">K</a> _(b))/(Delta T_(b) xx W_(1))= (1000 xx <a href="https://interviewquestions.tuteehub.com/tag/8-336412" style="font-weight:bold;" target="_blank" title="Click to know more about 8">8</a> xx 2.02)/(1.26 xx 100) = (161.6)/(1.26)` <br/> `= 128.25 g mol ^(-1).`</body></html>
Discussion
49398.

A solution containing 6 gm of a solute dissolved in 250 ml of water gave an osmotic pressure of 4.5 atmosphere at 27^(@)C. Calculate the boiling point of the solution. The molal elevation constant for water is 0.52

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> :[373.095]</body></html>
Discussion
49399.

A solution containing 2.68 xx 10^(-3)mol of A^(n+)ions requires 1.61xx10^(-3)mol of MnO_(4)^(-)for the complete oxidation of A^(n+) "to "AO_3^(-)in acidic medium. What is the value of n ?

Answer» <html><body><p><br/></p>Solution : <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 `A^(<a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a>+)` = Eqs.of `MnO_4^(-)` <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>.68xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)xx(5-n)=1.61xx10^(-3)<a href="https://interviewquestions.tuteehub.com/tag/xx5-2340058" style="font-weight:bold;" target="_blank" title="Click to know more about XX5">XX5</a>, :. N ~~2`</body></html>
Discussion
49400.

A solution containing 2.5 g of a non-volatile solute in 100 gm of benzene boiled at a temperature 0.42K higher than at the pure solvent boiled. What is the molecular weight of the solute? The molal elevation constant of benzene is "2.67 K kg 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> :`K_(b)="<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>.67 K <a href="https://interviewquestions.tuteehub.com/tag/kg-1063886" style="font-weight:bold;" target="_blank" title="Click to know more about KG">KG</a> mol"^(-1)` <br/> `DeltaT_(b)=0.42K` <br/> `W_(1)=100g=(100)/(1000)kg=0.1Kg` <br/> `W_(2)=(K_(b))/(DeltaT_(b)).(W_(2))/(W_(1))` <br/> `M_(2)=(2.67)/(0.42)<a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a>(2.5)/(0.1)` <br/> `M_(2)="158.98 g mol"^(-1)`</body></html>
Discussion