52671 + Interview Questions in Chemistry in Class 11

1.	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» SOLUTION :`{:(,.^(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",[0.209/57-10^(-5)],[0.6/56-10^(-5)],10^(-5),10^(-5)):}` `.^(57)Fe^(2+)` left after reaction `= 3.667xx10^(-3) - 10^(-5)` `= 366.6xx10^(5)` mole or `N_(0) = 366.6xx10^(-5)xx6.023xx10^(23) = 2.208xx10^(21)` Also `t = (2.303)/(lambda) log (N_(0))/(N)` `1 = (2.303xx4.62)/(0.693) log (2.208xx10^(21))/(N)` `N = 1.9xx10^(21)` `:.` Rate of decay `= lambda xx N = (0.693)/(4.62) xx 1.9xx10^(21)` `= 2.85xx10^(20)` DPH

Discussion

2.	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» 13.33 ML 6.66 ml 3.33 ml 26.67 ml Solution :No. of MILLI equivalents of `KMnO_4` = No of milli equivalents of THIOSULPHATE ions `V xx0.1 xx3=(0.158)/(((158)/8))xx1000` `V = 80/3 =26.67 ml`

Discussion

3.	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» Solution :` K_(SP)=[CD^(+2)][OH^(-) ] ^(2), 2 XX 10 ^(-14)=10 ^(-2)[OH^(-)]^(2) ` ` [OH^(-) ] =1.414 xx 10 ^(-6)rArr K_b =([NH_4^(+) ][OH^(-) ])/( [NH_4OH]) ` ` 1.8 xx 10 ^(-5)=([ NH_4^(+)][1.4 xx 10 ^(-6)])/( 10^(-2))` ` [NH_4^(+) ]=1.2 xx 10 ^(-1) `

Discussion

4.	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» CARBANION carbine for radical carbocation Answer :D

Discussion

5.	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» ` 1.3 XX 10^(-3) M` ` 2.6 xx 10^(-2)M` ` 3.7 xx 10^(-2) M` ` 5.8 xx 10^(-7) M` Solution :`[Sr^(+2) ][CO_3^(2-)] =KsP_1 ,[Sr^(+2) ][F^(-) ] ^(2) = Ksp_2` `rArr [F^(-) ]^(2) =(Ksp_2)/(Ksp_1) xx [CO_3^(2-)]` ` [F^(-)]^(2)=(8xx 10^(-10))/( 7 xx 10 ^(-10)) xx 1.2 xx 10 ^(-3)` ` [F^(-)] =3.7 xx 10 ^(-2)`

Discussion

6.	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» ` 1.3 xx 10^(-3) M` ` 2.6 xx 10^(-2)` ` 3.7 xx 10^(-2) M` ` 5.8 xx 10^(-7) M` Solution :`[Sr^(+2) ][CO_3^(2-)] =KsP_1 ,[Sr^(+2) ][F^(-) ] ^(2) = Ksp_2` `RARR [F^(-) ]^(2) =(Ksp_2)/(Ksp_1) xx [CO_3^(2-)]` ` [F^(-)]^(2)=(8XX 10^(-10))/( 7 xx 10 ^(-10)) xx 1.2 xx 10 ^(-3)` ` [F^(-)] =3.7 xx 10 ^(-2)`

Discussion

7.	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» `9.0` `8.0` 8.5979 5.4021 Answer :C

Discussion

8.	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» `46/90` `90/46` `46/(90 + 46)` `1/6` SOLUTION :No. of moles `(n_(A))` of ethyl alcohol = `46/46=1` No. of moles `(n_(B))` of WATER `=90/18=5` `therefore` Mole FRACTIONI of ethyl alcohol `=n_(A)/(n_(A) + n_(B)) = 1/(1+5) = 1/6`

Discussion

9.	A solution is prepared by dissolving 18.25 g NaOH in 200 mL of it. Calculate the molarity of the solution.
Answer» Solution :MOLARITY of solution (M) `=(("Mass of NaOH")/("MOLECULAR mass of NaOH"))/("Volume of solution in mL"/(1000))=((18.25g))/(("40 g mol"^(-1))xx("0.2 L"))` `= 2.281 "mol L"^(-1) = 2.281 M`.

Discussion

10.	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» <P> SOLUTION :(i)`x_A =( n_A)/( n_A+n_B)` ` n_A ="No of moles WATER "` `n_B = "No. of MOLESOF glucose " ` `n_A =( 100)/( 18 ) = 5.555` ` n_B =(2)/( 180) = 0.0111` Total moles ` (n_A+n_B) = 5.566` (ii)Urea ` n_A = "No. of moles of water " ` ` n_B=" No. of mole of urea " ` `n_A = 5.555` ` n_B=( 4)/( 60) = 0.0666 " moles of urea "` Total moles ` = 0.555+ 0.066 moles ` `x_B = ( n_A)/( n_A+n_B) ` `= ( 5.555)/( 5.621) = 0.9882` `P_B = P^(@) xx x_B ` `""= 23.756xx 0.9882` ` P_B = 23.42 torr`

Discussion

11.	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» Solution :Molecular mass of oxalic acid, `(COOH)_(2) . 2H_(2)O = 126` BASICITY of oxalic acid =2 `therefore` EQUIVALENT mass of oxalic acid `=126/2 = 63` Number of gram equivalents of oxalic acid DISSOLVED `=0.63/63 = 0.01` Volume of the solution `=100 cm^(3) = 100/1000 L = 0.1 L` NORMALITY of the solution `=("No. of g eq. of oxalic acid")/("Volume in litres")` `=0.01/0.1 = 0.1 g eq. L^(-1)`

Discussion

12.	A solution is prepared by adding 4g of a solute A to 36 of water. Calculate the mass percent of the solute.
Answer» ANSWER :MASS percent `=(4)/(40) xx 100=10%`

Discussion

13.	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» 0.1 0.111 0.9 0.75 SOLUTION :Mass PERCENT of X `=("Mass of X")/("Mass of solution")XX100` `5/(5+45)xx100`=10%

Discussion

14.	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» Solution :Masspercent of A `= ("Mass of A")/("Mass of solution") XX 100` `= (2G)/(2 g A + 18 g "of water") xx 100` `= (2g)/(20g) xx 100 = 10%`

Discussion

15.	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» SOLUTION :Mass PER cent of `A="Mass ofA"/"Mass of solution"xx100=(2g)/("2g of A+18g of WATER")xx100=(2g)/(20g)xx100=10%`

Discussion

16.	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» Solution :300 g of `25%` solution containsd solute = 75 g 400 g of `40%` solution contains solute = 160 g TOTAL MASS of soute `= 160 + 75 = 235g` Total mass of solution `= 300 + 400 = 700 g` `%` of solute in the final solution `= (235)/(700) xx 100 = 33.5%` `%` of water in the final solution `= 100 - 33.5 = 66.5%`

Discussion

17.	A solution is made by dissolving 49g of H_2SO_4 in 250 mL of water. The molarity of the solution prepared is
Answer» 2 M 1 M 4 M 5 M Solution :Molarity `=("WT. of solute")/("Mol. Wt. of solute")xx(1000)/("Volume of soln. (mL)")` =`49/98xx1000/250` =2M

Discussion

18.	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» 2 M 1 M 4 M 5 M Solution :Molarity (M) `= ("Weight of SOLUTE")/("Molecular weight of solute")xx(1000)/("Volume of solution (ml))=(49)/(98)xx(1000)/(250)=2M`

Discussion

19.	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» Solution :Concentrationof `HNO_(3)` solution = 0.63 g PER 100 ML (Given) `=6.3 g ` per litre `= (6.3)/(63) ` moles/litre = `10^(-1)M` (`:' ` Mol. Mass of `HNO_(3) = 63`) Now, `HNO_(3)`, completely ionizes as : `HNO_(3)H_(2)O rarrH_(3)O^(+)+NO_(3)^(-)` `:. [H_(3)O^(+)]=[HNO_(3)]=10^(-1)M :. pH = - log [H_(3)O^(+)]= - log 10^(-1)=1`

Discussion

20.	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» ANSWER :`126`

Discussion

21.	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» Solution :`POH = pK_(B) + log .(["SALT"])/(["BASE"])=4.75 + log.(0.2)/(0.1)` `=4.75+ log 2 = 4.75 + 0.3010 ~=5.0` `:. PH = 14 - 5.0 = 9`.

Discussion

22.	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» ` 3.5 xx 10^(-7)` ` 6.1 xx 10^(-8) ` ` 2.2 xx 10^(-7) ` ` 8.3 xx 10^(-8) ` SOLUTION :` K_(sp)=[AG^(+) ][Cl^(-) ] ` ` [Ag^(+) ] =(1.8 xx 10^(-10))/( 10^(-1) )= 1.8 xx 10^(-9) ` ` [I^(-) ]=(K_(sp) )/( [Ag^(+) ]_(cl^(-) ))=(1.5 xx10^(-16))/( 1.8 xx 10^(-19)) = 8.3 xx 10 ^(-8) M`

Discussion

23.	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» SOLUTION :MOLES of `CH_(3)OH=("Mass of "CH_(3)OH)/("MOLAR mass")=((60 G)/("32 g mol"^(-1)))=1.875` mol Moles of water `= ("Mass of water")/("Molar mass")=((120g)/("18 g mol"^(-1)))=6.667` mol Mole fraction of `CH_(3)OH=(("1.875 mol"))/(("1.875 mol +6.667 mol"))=(("1.875 mol"))/(("8.542 mol"))=0.220` Mole fraction water `= 1 - 0.220 = 0.780`.

Discussion

24.	A solution gives yellow colour with orange, methyl red and phenol red. What is the pH of the solution ?
Answer» SOLUTION :Yellow colour with methyl ORANGE MEANS `PH gt 4.5` Yellow colour with methyl redmeans pH ` gt6.2` Yellow colour with phenol red means `pH lt 6.4` . Hence, thesolution has pH between 6.2 to 6.4.

Discussion

25.	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» `7.4 XX 10^(-7)` `2.0 xx 10^(-7)` `6.1 xx 10^(-7)` `3.4 xx 10^(-7)` Solution :`K_(SP)(SrCrO_(4)) = [Sr^(2+)][CrO_(4)^(2-)]` `[CrO_(4)^(2-)] = (3.5 xx 10^(-5))/(0.1)= 3.5 xx 10^(-4)` `K_(sp)(BaCrO_(4)) = [Ba^(2)][CrO_(4)^(2-)]` `[CrO_(4)^(2-)]_("total") = [CrO_(4)^(2-)]` from `SrCrO_(4)` `[Ba^(2+)] = (1.2 xx 10^(-10))/(3.5 xx 10^(-4)) = 3.4 xx 10^(-7)`

Discussion

26.	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» `AgI` PRECIPITATES first The maximum `[I^(-)]` which can be maintained in the solution so that only `AgI` is precipitate is `10^(-6) M`. When the precipitation of `AGCL` just starts, `[I^(-)]` in the solution is `10^(-8) M` If sufficient NaIis added to a saturated solutions of `AgCl`, precipitate of `Agl` is obtained. Solution :`{:(10ml(N)/(20)NAOH +,20ml(N)/(20)HCl),("milli eq."=10 xx 1/20 = 1/2,"milli eq." = 20 xx 1/20 = 1):}` Solution also turn s methly orange red `[H^(+)] gt [H^(-)]` then (B) (C) and (D) are CORRECT.

Discussion

27.	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» 5.3 G and 4.2 g 3.3 g and 6.2 g 4.2 g and 5.3 g 6.2 g and 3.3 g Solution :N//A

Discussion

28.	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» SOLUTION :2.5 mL of `0.1 M H_(2)SO_(4)=2.5 mL " of" 0.2 N H_(2)SO_(4)` `=(1)/(2)Na_(2)CO_(3)` present in 10 mL of mixture So, `5 mL " of" 0.2 N H_(2)SO_(4)=Na_(2)CO_(3)` present in 10 mL of mixture `-=5 mL " of" 0.2 N Na_(2)CO_(3)` `-=(0.2xx53)/(1000)xx5=0.053 g` Amount of `Na_(2)CO_(3)=(0.053)/(10)xx1000=5.3 g//L` of mixture Between first and second end POINTS, =2.5 mL of `0.2 M H_(2)SO_(4)` used =2.5 mL of `0.4N H_(2)SO_(4)` used =5 mL of `0.2 N H_(2)SO_(4)` used `-=(1)/(2)Na_(2)CO_(3)+NaHCO_(3)` present in 10 mL of mixture `(5-2.5)mL 0.2 N H_(2)SO_(4)` `-=NaHCO_(3)` present in 10 mL of mixture `-=2.5 mL 0.2 N NaHCO_(3)` `-=(0.2xx84)/(1000)xx2.5=0.042 g` Amount of `NaHCO_(3)=(0.042)(10)xx1000=4.20 g//L` of mixture.

Discussion

29.	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» 5.3 G and 4.2 g 3.3 g and 6.2 g 4.2 g and 5.3 g 6.2 g and 3.3 g Solution :N//A

Discussion

30.	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» ` 0.1, 0.1` `0.1, 0. 05` ` 0.05, 0.1` ` 0.05, 0.05` Solution :Let molarity of `H_(2)SO_(4)= M_1 and H_2C_2O_4=M_2` No. of milli equivalents of `(H_2SO_4+H_2C_2O_4)` = No. of milli EQUIVALENT of NaOH `20(2M_1+2M_2)=40xx1/10` `40(M_1+M_2)=4` `M_1+M_2=0.1 ""...(1)` No.of milli equivalents of `H_(2)C_(2)O_(4)=` No. of mili equivalents `KMnO_4` `20(2M_2)=20 xx1/10` `M_(2) =1/20 =0.05 M...(2)` `M_1 =0.1 -0.05 =0.05M`

Discussion

31.	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» 9 5 7 `5.301` ANSWER :B

Discussion

32.	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» ANSWER :(a) `3.3669 CI, 0.6331 Ci`, (B) `0.3245`

Discussion

33.	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» ANSWER : 5.028 m, 4.187 M

Discussion

34.	A solution contains 25% water, 25% ethanol and 50% acetic acid by mass. Calculate the mole fraction of each component.
Answer» SOLUTION :Suppose, we have 100 g of the GIVEN solution. Then, mass of WATER = 25g mass of ethanol = 25 g and mass of acetic acid = 50 g. The molecular MASSES of water, ethanol and acetic acid are 18, 46 and 60 RESPECTIVELY. Therefore, number of moles of water `(n_(1)) = 25/18 = 1.39` number of moles of ethanol `(n_(2)) = 25/46 = 0.54` and number of moles of acetic acid `(n_(3)) = 50/60 = 0.83` Total number of moles in solution = `n_(1) + n_(2) + n_(3) = 1.39 + 0.54 + 0.83 = 2.76` Mole fraction of water = `n_(1)/(n_(1) + n_(2)+ n_(3))` `=1.39/2.76 = 0.503` Mole fraction of ethanol `=n_(2)/(n_(1) + n_(2) + n_(3))` `=0.54/2.76 = 0.196` Mole fraction of acetic acid = `n_(3)/(n_(1) + n_(2) + n_(3)) = 0.83/2.76 = 0.301`

Discussion

35.	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» Solution :Density of NACL solution `= 1.25 "G cm"^(-3)` , Volume of solution `= 95 cm^(3)` MASS of solution = volume `xx` density `= ("95 cm"^(3))xx("1.25 g cm"^(-3))=118.75`g Mass PERCENT of NaCl `= ("Mass of NaCl")/("Mass of solution")xx100` `=((20g))/((118.75g))xx100=16.84%`.

Discussion

36.	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» ANSWER :(a) `3.33xx10^(-2)M`, (B) `88.8xx10^(7) cpm//mL`

Discussion

37.	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» Solution :This reaction is an example of electrophilic AROMATIC substitution. In alkaline MEDIUM, phenol generates phenoxide ION which is more electron rich than phenol and hence more reactive for electrophilic ATTACK. 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.

Discussion

38.	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» SOLUTION :First ionisation, step 1 `,H_2 ShArr H^(+) +HS^(-) """ K_a = 1 xx 10^(-7)` Second ionisation, step 2,`HS^(-)hArr H^(+)+s^(2-) "" K_(a_2) = 3 xx 10^(-13)` Presence of HCI supresses the dissociation of `H_2 S` due to common ion effect of `H^(+)`. For step `1,1 xx 10^(-7)= ([H^+][HS^(-)])/( [H_2 S])=(0.3[HS^(-)])/( 0.1)` Cocentrationof `HS^(-) `,`[HS^(-)])= 3.3 xx 10^(-8)mol L^(-1)` or step` 2,1,.3 xx 10^(-13) = ([H^+] [S^(2-)])/([HS^(-)]) = ( 0.3 [S^(2-)])/(3.3 xx 10^(-8))` concentrationof `s^(2-) , [S^(2-)] = 1.43 xx 10^(-20) mol L^(-1)`

Discussion

39.	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» `Cl^(-) ` ions are precipitated first `CrO_4^(2-)` ions are precipitated first Both `Cl^(-) and CrO_4^(2-)` ions are precipitated TOGETHER The second ion starts PRECIPITATING when ` [1^(st) ion ] =2.758 xx 10^(-5)` Solution :` (a) [Ag^(+) ] _(cl^(-)) = ( 1.7 xx 10^(_10) )/(5xx 10 ^(_2)) =3.4 xx 10 ^(_9) M` ` [Ag^(+) ]_(CrO_4^(_2) ) = sqrt( (1.9 xx 10^(-12))/(5 xx 10 ^(_2)))=6.1 xx 10 ^(_6) M` ` Cl^(_) `Starts precipitation first ` [Cl^(-)] =(K_(SP))/( [Ag^(+)]_(CrO_4^(_2)))= (1.7 xx 10 ^(-10))/(6.1 xx 10^(_6)) ` ` ""= 2.78 xx 10 ^(_5) M`

Discussion

40.	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» `Mg(OH)_(2) (K_(SP)=3xx10^(-11))` `Fe(OH)_(2) (K_(sp)=8xx10^(-16))` `Cd(OH)_(2) (K_(sp)=8xx10^(-6))` `AgOH(K_(sp)=5xx10^(-3))` Solution :When equal volumes of `(NH_(4)Cl+NH_(4)OH)` and metal ions are mixed (volume becomes double and concentration is halved) `:. [OH]=(10^(6))/(2),[M^(+N)]=(0.1)/(2)` `Q_(sp)` ( or `IP`) of metal hydroxides of `M(OH)_(2)` type `=[M^(+n)][OH]^(2)` `=((0.1)/(2))((10^(-6))/(2))^(2)=(1)/(8)xx10^(-13)` `=0.125xx10^(-13)=12.5xx10^(-11)` `:. Q_(sp) lt K_(sp)` of `Mg(OH)_(2) (1.25xx10^(-11)gt3xx10^(-11))` and `Q_(sp)gtK_(sp)` of `Fe(OH)_(2) (1.25xx10^(-11)gt8xx10^(-16))` 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 FIRST. Similarly, `Q_(sp)` (or `IP`) of metal hydroxides of `M(OH)` type `=[M^(+1)][OH]=((0.1)/(2))((10^(-6))/(2))=0.25xx10^(-7)` `= Q_(sp)ltK_(sp)` of `AgOH(0.25xx10^(-7)lt5xx10^(-3))` It cannot be precipitated out. Hence `Fe(OH)_(2)` will be precipitated

Discussion

41.	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» 0.5 G 1g 2g 4g Solution :300 ml of 0.1 N HCl will neutralize all the NAOH and HALF 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 USED 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.

Discussion

42.	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» 0.5 g 1 g 2 g 4 g Solution :300 mL HCl of 0.1 N NEUTRALISES 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. THUS, 250 mL of 0.1 N HCl is required to neutralise NaOH completely. `N_(1)V_(1)(NaOH)=N_(2)V_(2)(HCl)` `=0.1xx250` =25 `W_(NaOH)=(ENV)/(1000)=(40xx25)/(1000)=1 g`

Discussion

43.	A solution containing Fe^(2+) ions is titrated with KMnO_(4) solution, Indicator used will be :
Answer» phenolphthalein methyl orange litmus none of these Solution :N//A

Discussion

44.	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» The difference of the number of m " mol of "`Cu^(2+)` and `C_(2)O_(4)^(2-)` ions in the SOLUTION is 10 m mol 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. The equivalent weight of `Cu^(2+)` ions in the titration with KI is equal to the atomic weight of `Cu^(2+)` The equivalent weight of KI in the titratio is `(M)/(2)` `(M=` molesular weight of KI) Solution :(i). `Cu^(2+)` does not react with `MnO_(4)^(ɵ)` Only `C_(2)O_(4)^(2-)` reacts with `MnO_(4)^(2-)` `MnO_(4)^(ɵ)=C_(2)O_(4)^(2-)` `(n=5)(n=2)` `mEq-=mEq` `20xx(M)/(4)xx5-=mEq` `thereforem" Eq of "C_(2)O_(4)^(2-)=25` `mmoles of `C_(2)O_(4)^(2-)=(25)/(2)=12.5` (II). `Cu^(2+)=KI-=I_(2)=S_(2)O_(3)^(2-)` (hypo) `mEq-=mEq-=mEq-=mEq` `(2Cu^(2+)+2I^(ɵ)toCu_(2)I_(2))(2S_(2)O_(3)^(2-)to_S_(4)O_(6)^(2-)+2e^(-))` `(e+Cu^(2+)toCu^(1+))` `m" Eq of "S_(2)O_(3)^(2-)-=25xx(M)/(10)xx1` `=2.5m" Eq of "Cu^(2+)` `m" Eq of "Cu^(2+)=2.5` `mmoles of Cu^(2+)=(2.5)/(1)=2.5` Difference in mmoles of `C_(2)O_(4)^(2-)` and `Cu^(2+)=12.5-2.5=10` EW of `Cu^(2+)=("atomic weight" of Cu^(2+))/(1("n-factor"=1))` Ew of `KI=(M)/("nfactor")=(M)/(1)=M{:(2I^(ɵ)toI_(2)+2e^(-)),(n=(2)/(2)=1):}`

Discussion

45.	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» `8` ` 9` ` 7` `6` Solution :` H_3 PO_4 + NaOH to NaH_2PO_4 +H_2O ` ` {:( 10 m " MOLES " , 15 m " moles" ,0) , ( - ,10" moles " , 10 "moles " ), (- , 5 " moles " , 10 m " moles ") :} ` ` NaH_2PO_4 + NaOH to Na_2HPO_4 + H_2O ` ` {:( 10 m " moles" , 5 m " moles ",0),( 5, - , 5 m " moles " ),( 5 m " moles" ,- , 5 m " moles") :} ` ` pH =P^(K_(A2)) + log ""(S)/(A)rArr pH = 8`

Discussion

46.	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» 50 ml 150 ml 200 ml 250ml SOLUTION :In the above question if we USED 200 ml, 0.1 MNaOH, only SALT`Na_2HPO_4` is formed

Discussion

47.	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» SOLUTION :We have, mass of SOLUTE,` W_(2)=8g` Mass of solvent, `W_(1) = 100 g ` Elevation of boiling point, `Delta T_(b) = 36.86 - 35.60=1.26 ^(@)C` `K_(b)=2.02` Molecular mass of the solute `M _(2) = ( 1000 xx W_(2) xx K _(b))/(Delta T_(b) xx W_(1))= (1000 xx 8 xx 2.02)/(1.26 xx 100) = (161.6)/(1.26)` `= 128.25 g mol ^(-1).`

Discussion

48.	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» ANSWER :[373.095]

Discussion

49.	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» Solution : EQUIVALENTS of `A^(N+)` = Eqs.of `MnO_4^(-)` `2.68xx10^(-3)xx(5-n)=1.61xx10^(-3)XX5, :. N ~~2`

Discussion

50.	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» SOLUTION :`K_(b)="2.67 K KG mol"^(-1)` `DeltaT_(b)=0.42K` `W_(1)=100g=(100)/(1000)kg=0.1Kg` `W_(2)=(K_(b))/(DeltaT_(b)).(W_(2))/(W_(1))` `M_(2)=(2.67)/(0.42)XX(2.5)/(0.1)` `M_(2)="158.98 g mol"^(-1)`

Discussion

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