94420 + Interview Questions in Chemistry in Class 12

1.	An aqueous solution of a mixture of two inorganic saits, when treated with dilute HCl. gave a precipilate (P) and a filtrate (Q). The precipitate P was found to dissolve in hot water. The filtrate (Q) remained unchanged, when treated with H_(2)S in a dilute mineral acid medium. However, it gave a precipitate (R) with H_(2)S in an ammoniacal medium. The precipitate R gave a coloured solution (S), when treated with H_(2)O_(2) in an aqueous NaOH medium. The precipitate P contains
Answer» `Pb^(2+)` `Hg_(2)^(2+)` `AG^(+)` `Hg^(2+)` Solution :`{:(Pb^(2+)+Cr^(3+) OVERSET("DIL. HCl")(rarr)PbCl_(2) DARR +CrCl_(3)(Q)),("WHITE ppt."),("soluble in hot"),("water insoluble in"),("cold"):}`

Discussion

2.	An aqueous solution of a metal nitrate yields a white precipitate on treatment with aqueous Na_(2)SO_(3) solution. Precipitate dissolves in NaOH(aq) solution forming a complex anion X. Also, the orginal sample of solution yields a white precipitate on treatment with dilute HCl solution, which dissolves in boiling water. The most likely formula of the complex anion (X) is
Answer» `[Ag(OH)_(2)]^(-)` `[Ag(OH)_(4)]^(-)` `[ZN(OH)_(4)]^(2-)` `[Pb(OH)_(4)]^(2-)` Answer :D

Discussion

3.	An aqueous solution of a gas (X) turns red litmus blue. When added in excess to copper sulphate solution, deep blue colour is obtained. On addition to a ferric .chloride solution a brown precipitate soluble in dilute nitric acid is obtained. Hence, (X) is
Answer» `SO_(2)(G)` `H_(2)S(g)` `NO_(2)(g)` `NH_(3)(g)` Solution :The gas (X) is `NH_(3)`. `NH_(3)(g)+"Red litmus" to "Blue"` `CuSO_(4)+underset((X))(4NH_(3))(aq)tounderset("Deep blue")([CU(NH_(3))_(4)])SO_(4)` `3NH_(3)+3H_(2)O+FeCl_(3)toFe(OH)_(3)+3NH_(4)CL` `Fe(OH)_(3)+3HNO_(3)toFe(NO_(3))_(3)+3H_(2)O` Hence,(D) is the corret answer.

Discussion

4.	An aqueous solution of a metal bromide MBr_(2)(0.05M) is saturated with H_(2)S. What is the minimum pH at which MS will precipitate ? K_(SP) for M S= 6.0xx10^(-21) . Concentration of saturqated H_(2)S=0.1M, K_(1)=10^(-7)and K_(2)=1.3xx10^(-13) for H_(2)S .
Answer» ANSWER :0.983

Discussion

5.	An aqueous solution of a metal bromide MBr_(2)(0.05M) is saturated with H_(2)S. What is the minimum pH at which MS will precipitate ? K_("sp") for Ms = 6.0xx10^(-21) Concentration of saturated H_(2)S=0.1 M, K_(1)=10^(-7) and K_(2)=1.3xx10^(-13)" for "H_(2)S. [Report your answer by rounding it upto nearest whole number]
Answer» ANSWER :`1.00`

Discussion

6.	An aqueous solution of a gas (X) gives the following reactions: On passing H2S into the solution, turbidity is obtained. Identify (X) and give equations for the steps (i), (ii), (iii) .
Answer» SOLUTION :`X-SO_(2)` `SO_(2)+2H_(2)S rarr underset("TURBIDITY")underset("WHITE")(3S darr +2H_(2)O)`

Discussion

7.	An aqueous solution of a gas (X) gives the following reactions: It decolourizes an acidified K_(2)Cr_(2)O_(7) solution.
Answer» Solution :`X-SO_(2)` `K_(2)Cr_(2)O_(7)+H_(2)SO_(4)+3SO_(2) rarr K_(2)SO_(4)+Cr_(2)(SO_(3))_(3)+H_(2)O`

Discussion

8.	An aqueous solution of a compound (A) is acidic towards litmus and the solid has the property to sublime at 300^(@)C. The compound (A) on treatement with an excess of NH_(4)SCN gives a red coloured compound (B) an on treatement with a solution of K_(4)[Fe(CN)_(6)] gives a blue coloured compound (C ) . The compound (A) on heating with excess of K_(2) Cr_(2) O_(7) and concentrated H_(2)SO_(4) evolves deep red vapours of (D) . On passing these vapours through NaOH solution and then adding acetic acid and lead acetate solutions, a yellow precipitate of compound (E ) is obtained . Identifycompounds ( A) to (E ) and give balanced chemical equation for each reaction involved.
Answer» Solution :`FeCl_(3)` solution is acidic and the solid sublimes at `300^(@)C`. Its solution reacts with `NH_(4)SCN` to give red coloured ferric thiocyanateand react with `K_(4)FE(CN)_(6)` to FORM a blue coloured complex, ferric ferrocyanide `underset((A))(FeCl_(3)) + underset( "Ammonium thiocyanate")(3NH_(4)SCN)rarr underset("Ferric thiocyanate ( Red)") underset((B))(Fe(SCN)_(3))+ 3NH_(4)Cl` `4FeCl_(3)+ 3K_(4)[Fe(CN)_(6)] rarr underset("Ferric ferrocyanide (Blue)") underset((C )) (Fe_(4)[Fe(CN)_(6)]_(3))= 12 KCl` `underset((A)) (2FeCl_(3))+ K_(2)Cr_(2)O_(7)+ 3H_(2)SO_(4) rarr underset("Chromyl chloride (Red vapours)")underset((D))(2CrO_(2)Cl_(2)) + Fe_(2)(SO_(4))_(3) +2KCl+ 3H_(2)O` `CrO_(2)Cl_(2)+ 4NaOH rarr underset("Sodium chromate")(Na_(2)CrO_(4)) + 2NaCl+ 2H_(2)O` `Na_(2)CrO_(4)+ (CH_(3)COO)_(2)Pboverset(CH_(3)COOH)(rarr) underset("Lead chromate (yellow ppt.)") underset((E )) (PbCrO_(4)) + 2CH_(3)COONA`

Discussion

9.	An aqueous solution of a dibasic acid (mol. Mass = 118) containing 35.5 g of the acid per litre of the solution has density 1.0077g//cm^(3). Express the concentration of the solution in as many ways ways as you can.
Answer» Solution :`"Molarity "=(35.4)/(118)"mol L"^(-1)="0.3M,Noramality "=(35.4)/(118//2)"G EQ L"^(-1)=0.6N` `1000CM^(3) " solution "=1000xx1.0077 g = 1007.7 g` `"Solvent (WATER) "=1007.7 - 35.4=972.3 g = 0.9723 kg` `"Molality "=("35.4/118 mol")/("0.9723 kg")=0.3085 ~=0.31 , x_(2)=(n_(2))/(n_(1)+n_(2))=(35.4//118)/(972.3//18+35.4//118)=0.0055` `x_(1)=1-x_(2)=0.9945.`

Discussion

10.	An aqueous solution of a gas (G) decolourises acidified permanganate and pararosanilide. On boiling G with hydrogen peroxide, an acid X is formed, which gives a white precipitate Y with barium chloride solution. Y is insoluble in dilute hydrochloric acid. Identify the compounds G and Y.
Answer» SOLUTION :`G=SO_(2),Y=BaSO_(4)`

Discussion

11.	An aqueous solution of a compound (X) when treated with BaCl_(2) solution gives a white precipitate in soluble in concentrated HCl. Another sample of (X) gives first white precipitate with NaOH which is soluble in excess of NaOH solution. The solution of (X) does not give the precipitate on passing H_(2)S gas. Identify the compound (X) and give necessary reactions.
Answer» Solution :`Al_(2)(SO_(4))_(3)+3BaCl_(2)rarr2AlCl_(3)+underset("WHITE")(3BaSO_(4)darr)` `Al_(2)(SO_(4))_(3)+6NaOH rarr underset("white")(2Al(OH)_(3))+3Na_(2)SO_(4)` `AL(OH)_(3)+NaOHrarrunderset("SOLUBLE")(NaAlO_(2))+2H_(2)O` Compound (X) gives white precipitate with `BaCl_(2)` which is insoluble in concentrated HCl hence the anion of (X) MUST be `SO_94)^(2-)` ion. Because the cation gives white precipitate with NaOH which is soluble in excess of NaOH, therefore cation of compound (X) may be `ZN^(2+), ` or `Al^(3+)` ion. But `Zn^(2+)` cannot be cation of the compound (X).

Discussion

12.	An aqueous solution of 6.3 g of oxalic acid dihydrate is made upto 250 mL. The volume of 0.1 N NaOH required to completely neutralise 10 mL of this solution is
Answer» 40 ML 20 mL 10 mL 4 mL Solution :Normality of OXALIC acid solution `=6.3/250 XX 1000/63 = 0.4 N` `10 xx 0.4 = V xx 0.1` or V = 40 mL

Discussion

13.	An aqueous solution of 6.3 g of oxalic acid dihydrate is made upto 250 mL. The volume of 0.1 N NaOH required to completely neutralize 10 mL of this solution is
Answer» 40mL 20mL 10mL 5mL Solution :Normality of OXALIC acid ` = (6.3 xx 1000)/(63 xx 250)` = 0.4 N `N_1V_1 = N_2V_2` ` 10 xx 0.4 = 0.1 xx V` `V = (10 xx 0.4)/(0.1) = 40 mL`

Discussion

14.	An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normalboiling point of the solvent. What is the molar mass of the solute ?
Answer» Solution :Applying Raoult.s law of dilute solution `(p^0 - p_s)/(p^0) = (n_2)/(n_1+n_2) = n_2/n_1 = (w_2// M_2)/(w_1//M_1)= w_2/M_2 xx M_1/w_1` SUBSTITUTING the values, we have `((1.013 - 1.004)"BAR")/(1.013 "bar") = (2g)/(M_2) xx (18 G "mol"^(-1))/(98 g)` `M_2 = (2xx 18)/(98) xx (1.013)/(0.009) g "mol"^(-1) =41.g "mol"^(-1)`

Discussion

15.	An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute ?
Answer» Solution :Vapour pressure of PURE water at the boiling POINT = 1 atm = 1.013 BAR Vapour pressure of solution=1.004 bar Mass of solute = 2 g Mass of solution = 100 g Mass of solvent = 100 - 2 = 98 g Applying Raoult.s LAW, `(p^@ -p_S)/(p^@) = X_2 ~= (w_2 xx m_1) /(m_2 xx W_1) ` `(1.013 -1.004 )/(1.013 ) =(2xx 10 )/(M_2 xx 98 )` `orM_2 =(2xx 18xx 1.013 )/(0.009xx 98 )`

Discussion

16.	An aqueous solution of 2 % non - volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solven. What is the molar mass of the solute ?
Answer» <P> Solution :Here, Vapour pressure of the solution at normal boiling point `(p_(1))=1.004` bar Vapour pressure of pure water at normal boiling point, `(p_(1)^(0))=1.013` bar Mass of SOLUTE, `(w_(2))=2 g` Mass of solvent (water), `(w_(1))=98 g` MOLAR mass of solvent (water), `(M_(1))=18 g mol^(-1)` According to Raoult.s LAW, `(p_(1)^(0)-p_(1))/(p_(1)^(0))=(w_(2)xxM_(1))/(M_(2)xx w_(1))` `(1.013-1.004)/(1.013)=(2xx18)/(M_(2)xx98)` `M_(2)=(1.013xx2xx18)/(0.009xx98)` `= 41.35 g mol^(-1)` Hence, the molar mass of the solute is 41.35 g `mol^(-1)`.

Discussion

17.	An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molecular mass of the solute?
Answer» Solution :Vapour pressure of PURE water at the boiling point `(p^(@))=" 1 atm= 1.013 bar"` Vapour pressure of solution `(p_(s))="1.004 bar , MASS of solute "(w_(2))=2 g` `"Mass of solution = 100 g,Mass of solvent = 98 g"` APPLYING Raoult's law for dilute solution `"(being %)"=(p^(@)-p_(s))/(p^(@))-(n_(2))/(n_(1))~=(n_(2))/(n_(1))=(w_(2)//M_(2))/(w_(1)//M_(1))=(w_(2))/(M_(2))xx(M_(1))/(w_(1))` `THEREFORE((1.013-1.004))/("1.013 bar")=(2g)/(M_(2))xx("18 g mol"^(-1))/("98 g")"or"M_(2)=(2xx18)/(98)xx(1.013)/(0.009)" g mol"^(-1)="41.35 g mol"^(-1)`

Discussion

18.	An aqueous solution of 2% non-volatile exerts a pressure of 1.004 Bar at the normal boiling point of the solvent. What is the molar mass of the solute ?
Answer» <P> SOLUTION :`(P_(A)^(0) - P_(A))/(P_(A)^(0)) = (w_(B) XX m_(A))/(m_(B) xx w_(A))` `(1.013 - 1.004)/(1.013) = (2 xx 18)/( m_(B) xx 98)` `m_(B) = 41.35` gm/mol

Discussion

19.	An aqueous solution of 0.5 g of NaOH is dissolved in 500 cm^3 of the solution. The molarity of the solution will be :
Answer» 0.025M 0.25M 2.5M None of the above Answer :A

Discussion

20.	An aqueous solution of 0.02 M KCl solution is filled in a 25-cm-long capillary tubeof internal radius 0.01 cm. The solution was found to have a specific conductance of 0.0027 "mho cm"^(-1). What will be the current in amp when a potential of 2 volts is appled across the capillary tube ?
Answer» SOLUTION :`6.78 XX 10^(-8)` MP

Discussion

21.	An aqueous solution of 0.1 M FeCl_3 is isotonic to which of the following aqueous solutions ?
Answer» `0.4N BaCl_2` `0.4N Al_2(SO_4)_(3)` `0.4N Na_3PO_4` `0.4N NACL`

Discussion

22.	An aqueous solution of 0.1 M NH_4Cl will have a pH closer to:
Answer» 9.1 8.1 7.1 5.1 Answer :D

Discussion

23.	An aqueous solution of 0.01 gram of a polymer in 10ml of water at 293K shows a 5.22 cm rise in the level (density of solution =gram /c c) . If molecular weight of polymer is (4.67xx10^(y)) then y=?
Answer» ANSWER :C

Discussion

24.	An aqueous solution is heated until it begins to boil. The atmospheric pressure is 760 mm of Hg. The boiling temperature will be
Answer» `100^(@)C` `gt100^(@) C` `lt100^(@)C` None Solution :The b.pt.of solution is ALWAYS HIGHER than b.pt.of solvent (`100^(o) C `for water).

Discussion

25.	An aqueous solution freezes at - 0.372^@C. If K_f and K_b for water are 1.86K kg "mol"^(-1) and 0.53 K kg "mol"^(-1) respectively, the elevation in boiling point of same solution in K is:
Answer» 0.72 0.46 4.6 0.106 Answer :D

Discussion

26.	An aqueous solution freezes at -0.2^(@)C. What is the molality of the soluiton ? Determines also (i) elevation in the boiling point (ii) lowering of vapour pressure at 298 K, given that K_(f)=1.86^(@)"kg mol"^(-1),K_(b)=0.512^(@)" kg mol"^(-1) and vapour pressure of water at 298 K is 23.756 mm.
Answer» Solution :`"For (ii) :"Delta T_(f)=(1000K_(f)w_(2))/(w_(1)M_(2))=(1000K_(f)n_(2))/(n_(1)M_(1))`. Calculate `(n_(2))/(n_(1))`. Then APPLY `(Deltap)/(p^(@))=(n_(2))/(n_(1))." Calculate "Delta p.`

Discussion

27.	An aqueous solution freezes at 271.5 K. Determine its boiling point and vapourpressure at 298 K. The cryoscopic constant of water is 1.86^@, its ebullioscopic constant is 0.516o and the water vapour pressure at 298 K is 3168 Pa.
Answer» SOLUTION :373.42, 3124 PA

Discussion

28.	An aqueous solution freezes at -0.186^(@)C (K_(f)=1.86^(@) ,K_(b)=0.512^(@). What is the elevation in boiling point?
Answer» 0.186 0.512 0.86 0.0512. SOLUTION :For the same solution `(DeltaT_(B))/K_(b)=(DeltaT_(b))/K_(F)` of `DeltaT_(b)=(DeltaT_(f))/(T_(f))xxK_(b)` `=((0.186^(@)C))/1.86xx0.512.` =`0.0512^(@)C`

Discussion

29.	An aqueous solution freezes at - 0.186^@C (K_f = 1.860, K_b = 0.512^@). What is the elevation on boiling point?
Answer» 0.186 0.512 0.86 0.0512 Answer :D

Discussion

30.	An aqueous solution contianing 20% by weight of liquid X (mol.wt. =140) has a vapour pressure 160mm at 60^(@)C. Calculate the vapour pressure of pure liquid 'X' if the vapour pressure of water is 150mm at 60^(@)C.
Answer» ANSWER :`470.5mm`

Discussion

31.	An aqueous solution contains the ions as Hg^(2+),Pb^(2+) anc Cd^(2+). The addition of dilute HCl (6N) precipitates
Answer» `Hg_(2)Cl_(2)` only `pBcL_(2)` only `PbCl_(2)` and `HgCl_(2)` `Hg_(2)Cl_(2)` and `PbCl_(2)` SOLUTION :`PbCl_(2),Hg_(2)Cl_(2)` are INSOLUBLE in WATER

Discussion

32.	An aqueous solution freezes at -0.186^@C (K_f= 1.86^@: K_b=0.512^@). What is the elevation in boiling point:
Answer» 0.186 0.512 0.512/1.86 0.0512 Answer :D

Discussion

33.	An aqueous solution contains Hg^(2+),Hg_(2)^(2+),Pb^(2+) and Cd^(2+). The addition of HCl (6N) will precipitate
Answer» `Hg_(2)Cl_(2)` only `PbCl_(2)` only `PbCl_(2)` and `Hg_(2)Cl_(2)` `PbCl_(2) and HgCl_(2)` Solution :Only `PbCl_(2)` and `Hg_(2)Cl_(2)` will precipitate as `Pb^(2+)` and `Hg_(2)^(2+)` as FIRST group basic radicals and their solubility PRODUCT is less than the other radicals.

Discussion

34.	An aqueous solution contains Hg^(2+),Hg_(2)^(2+),Pb^(2+) and Cd^(2+) Out of these how many ionswillproduce white precipitate with dilute HCl ?
Answer» Solution :`Hg_(2)^(2+) ,PB^(2+) `giveswhite ppt of `Hg_(2)CI_(2) and PbCI_(2)` with DIL HCI`

Discussion

35.	An aqueous solution contains Hg^(2+), Hg_(2)^(2+), Pb^(2+) and Cd^(2+). The addition of HCI (6N) will precipitate:
Answer» `Hg_(2)Cl_(2)` only `PbCl_(2)` and `Hg_(2)Cl_(2)` `PbCl_(2)` only `PbCl_(2)` and `HgCl_(2)` ANSWER :B

Discussion

36.	An aqueous solution contains Hg^(2+), Hg_2^(2+), Pb^(2+)andSb^(3+).The addition of HCl (6N) will precipitate :
Answer» `Hg_2Cl_2` only `PbCl_2` only `PbCl_2` and `Hg_2Cl_2` `PbCl_2` and`HgCl_2` ANSWER :C

Discussion

37.	An aqueous solution contains Hg^(2+), Hg_2^(2+),Pb^(2+),Ag^(+),Bl^(3+) and Cd^(2+).Out of these, how many ions will produce white precipitate with dilute HCl ?
Answer» Solution :`Pb^(2+)+2Cl^(-)toPbCl_2darr(white), Hg_2^(2+)+2Cl^(-)toHg_2Cl_2darr`(white) `Ag^(+)+Cl^(-)toAgCldarr`(white) `K_(sp)` of chlorides of `Pb^(2+)` and `Hg_2^(2+)` is low as compared to `K_(sp)` of `Hg^(2+)` and `CD^(2+)` . CHLORIDE ion concentration provided by dilute H Cl is just enough toexceed the `K_(sp)` of `PbCl_2` and `Hg_2Cl_2` Thus `Pb^(2+)` and `Hg_2^(2+)` are precipitated as their chlorides

Discussion

38.	An aqueous solution contains Hg^(2+),Hg_(2)^(2+),Pb^(2+),Ag^(+),Bi_(3+) and Cd^(2+).Out of these, how many ions will produce white precipitate with dilule HCl ?
Answer» Solution :`Pb^(2+) + 2Cl^(-) to PbCl_(2) darr ("white") , Hg_(2)^(2+) + 2Cl^(-) to Hg_(2)Cl_(2) darr` (white) `Ag^(+) + Cl^(-) to AgCl darr` (white) `K_(sp)` of CHLORIDES of `Pb^(2+)` and `Hg_(2)^(2+)` is low as compared to `K_(sp)` of `Hg^(2+)` and `Cd^(2+)`. Chloride ion concentration provided by dilute `HCl` is just enough to EXCEED the `K_(sp)` of `PbCl_(2)` and `Hg_(2)Cl_(2)`. Thus `Pb^(2+)` and `Hg_(2)^(2+)` are PRECIPITATED as their chlorides.

Discussion

39.	An aqueous solution contains Hg^(2+),Hg_(2)^(2+),Pb^(2+),Ag^(+),Bi^(3+) and Cd^(2+). Out of these, how many ions will produce white precipitate with dilute HCl.
Answer» Solution :`Pb^(2+)+2Cl^(-) to PbCl_(2) darr`(WHITE) `Hg_(2)^(2+)+2Cl^(-) to Hg_(2)Cl_(2)darr`(white) `Ag^(+) to Cl^(-) to AgCl darr` (white) `K_(sp)` of chlorides of `Pb^(2+)` and `Hg_(2)^(2+)` is LOW as compared to `K_(sp)` of `Hg^(2+)` and `Cd^(2+)`. Chloride ion concentration provided by dilute HCL is just enough to exceed the `K_(sp)` of `PbCl_(2)` and `Hg_(2)Cl_(2)`. Thus `Pb^(2+)` and `Hg_(2)^(2+)` are precipitated as their chlorides.

Discussion

40.	An aqueous solution contains an unknown concentration of Ba^(2+) when 50 mL of a 1 M solution of Na_(2)SO_(4) is added, BaSO_(4) just begins to precipitate. The final volume is 500mL. The solubility product of BaSO_(4) is 1xx10^(-10). What is the original concentration of BA^(2+)
Answer» `2xx10^(-9)M` `1.1xx10^(-9)M` `1.0xx10^(-10)M` `5XX10^(-9)M` Solution :`Ba^(+2)+SO_(4)^(2-)hArrunderset(ppt)(BaCo_(4(s)))` FINAL CONC. Of `[SO_(4)^(2-)]=(MV_(1))/(V_(1)+V_(2))=(1xx50)/(500)=0.1M` Final conc. Of `[Ba^(+2)]when BaSO_(4)` start PRECIPITATING `K_(SP)=Q_(SP)=[Ba^(+2)][SO_(4)^(2-)]` `10^(-10)=[Ba^(+2)](0.1M)` `[Ba^(+2)]=10^(-9)M` initial conc. `[Ba^(+2)],` initial volume was `=500-50=450` ml `M_(1)=(M_(2)V_(2))/(V_(1))=(10^(-9)xx500)/(450)` `M_(1)=1.1xx10^(-9)M.`

Discussion

41.	An aqueous solution contains a substance which yields 4 xx 10^-3 mol liter^-1 ion of H_3O^+. If log 2 = 0.3010 the pH of the solution is:
Answer» 1.5 2.398 3 3.4 Answer :B

Discussion

42.	An aqueous solution contains 5% and 10% of urea and glucose respectively (by weight). If K_(f) for water is 1.86 the f.p. of the solution is
Answer» `3.03K` `3.03^(@)K` `-3.03^(@)C` `-3.03K` ANSWER :C

Discussion

43.	An aqueous solution contains 5% by weight of urea and 10% by weight of glucose. What will be its freezing point? (Molal depression constant of water is 1.86^(@)C).
Answer» Solution :`"Mass of solution = 100 g ,Mass of urea = 5 g"` `"Mass of GLUCOSE = 10"therefore "Mass of water "=100-(5+10)=85h` `"Molality of urea in the solution "=(5)/(60)xx(1000)/(85)=0.98m` `"Molality of glucose in the solution "=(10)/(180)xx(1000)/(85)=0.65m` Total `DeltaT_(f)=K_(f)(m_(1)+m_(2))=1.86(0.98+0.65)=3.03^(@)C` `therefore"Freezing point of the solution "=-3.03^(@)C.`

Discussion

44.	An aqueous solution contains 5% and 10% of urea and glucose respectively (by weight). Find freezing point of the solution . K_(f) for water =1.86
Answer» Solution :We have `DeltaT_(F)=K_(f)xx` molality `=K_(f)xx` no. of moles of solute PER `1000g` of solvent `=1.86xx((5)/(60)+(10)/(180))xx(1000)/(85)=3.03` `{{:("urea"=60),("glucose"=180):}}` Since PURE water freezes at `0^(@)C`, the f.p. of the solution will be `-3.03^(@)C`

Discussion

45.	An aqueous solution contains 4% (w/v) NaOH and 5.3% (w/v) Na_(2)CO_(3). 40 ml of this solution is titrated with 2.0 M-HCl solution, using phenolphthalein indiactor. The volume of HCl solution needed for end point is
Answer» 30 ML 40 ml 20 ml 35 ml ANSWER :A

Discussion

46.	An aqueous solution contains 3 g of urea dissolved in 50 g of water. If the mole fraction of ethanol in diluted solution . The vapour pressure of water at the same temperature is 2267.7 Nm^(-2).
Answer» SOLUTION :`"Moles of urea "(n_(B))=((3g))/((60" g mol"^(-1)))=0.05 mol` `"Moles of urea "(n_(A))=((50g))/((18" g mol"^(-1)))=2.778 mol` `"Mole fration of water "(x_(A))=(n_(A))/(n_(A)+n_(B))=((2.778 mol))/((2.778 mol+0.05 mol))=0.9823` `"Vapour pressure of solution"(P)=P_(A)^(@)x_(A)` `=(2267.7 Nm^(-2))XX(0.9823)xx(0.9823)2227.6 Nm^(-2).`

Discussion

47.	An aqueous solution contains 10% by weight by urea (60.00)and 5% by weight of glucose (180.00). What will be its freezing point ? K_(f) for water is 1.86.
Answer» ANSWER :`(-4.254^(@)C)`

Discussion

48.	An aqueous solution contains 10g of glucose (mol.wt. =180) per 0.5L . Assuming the solution to be ideal calculate osmatic pressure at 25^(@)C
Answer» Solution :Mole of GLUCOSE `=(10)/(180)=(1)/(18)`, `V=0.5L` `R=0.082` lit. ATM / K / mole `T=273+25=298K` We have, `p=(n)/(V) RT` `=(1//18)/(0.5)xx0.082xx298` `=2.715atm`

Discussion

49.	An aqueous solution contains 10% ammonia by mass and has a density of 0.99 g/cc. Calculate the hydroxyl and hydrogen-ion concentration in this solution.K_a (NH_4^+) = 5 xx 10^(-10)M.
Answer» SOLUTION :`1.08 XX 10^(-2), 9.28 xx 10^(-13) M`

Discussion

50.	An aqueous solution containing urea was found to have boiling point more than the normalboiling point of water (373.13 K). When the same solution was cooled it was found that its freezing point is less than the normal freezing point of water (273.13 K). Explain these observations.
Answer» SOLUTION : BOILING point : Boiling point of a liquid is the temperature at which its vapour pressure is equalto the atmospheric pressure. Water boils at 373.13 K because at this temperature, the vapour pressure of water is equal to the atmospheric pressure (1.013 bar). HOWEVER, the vapour prcesurc of a liquid decreases in the presence of non-volatile solute like urea. In order that the vapour pressure of the solution reaches the atmospheric pressure, we shall have to raise the temperaturebeyond the normal boiling temperature of pure water. Thus, boiling temperature of a solution is always higher than that of pure solvent. Freezing point : Freezing point of a substance is the temperature at which the SOLID phase is in dynamic equilibrium with the liquid phase. We may define the freezing point of a substance as the temperature at which the vapour pressure of the substance in liquid phase is equal to its vapour pressure in the solid phase. When a non-volatile solid is added to the solvent, its vapour pressure decreases and now it would become equal to the that of the solid phase at a LOWER temperature. Thus the freezing point of a solution of urea is lower than that of pure water.

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