CHEMISTRY
CHAPTER PRACTICE PROBLEMS
FIITJEE
Liquid Solution
Name: ______________________________________ Batch:
Date:
Ideal and Non ideal solution
1. Match Column – I with Column – II
Column – I(Solution)
Column – II(nature)
(A) Benzene + toluene
(P) Non – ideal solution
(B) Ethanol + water
(Q) Ideal solution
(C) Benzene + chloroform
(R) ΔH mix > 0
(D) Carbon tetrachloride + chloroform
(S) ΔH mixing = 0
[Ans : A : Q,S; B : R,P; C : P,R; D : Q,S]
2. Match Column – I with Column – II
Column – I(Solution)
Column – II(Properties)
(A)
CCl4 + CHCl3
(P)
Raoult’s law
(B)
Benzene + H2O
(Q)
(C)
PA ∝ XA
(R)
(D)
PA < PA0 x A
(S)
Nearly ideal solution
Non – ideal solution with negative
deviation
Non – ideal solution with positive
deviation
[Ans : A : Q; B : S; C : P; D : R]
Vapour Pressure
3. For a binary ideal liquid solution, the total pressure of the solution is given as:
(A) Ptotal = PA* + PA* − PB* XB
(B) Ptotal = PB* + PA* − PB* X A
*
B
(C) Ptotal = P
(
+ (P
*
B
*
A
−P
)
)X
A
*
B
(D) Ptotal = P
(
+ (P
*
B
)
−P )X
*
A
B
[Ans:- B]
4. An ideal solution was found to have a vapour pressure of 80 torr when the mole fraction of a non – volatile
solute was 0.2. what would be the vapour pressure of the pure solvent at the same temperature?
(A) 64 torr
(B) 80 torr
(C) 100 torr
(D) 400 torr
[Ans:- C]
5. The vapour pressure of an aqueous solution of sucrose at 373 K is found to be 750 mm Hg. The molality of
the solution at the same temperature will be:
(A) 0.26
(B) 0.73
(C) 0.74
(D) 0.039
[Ans:- C]
6. An ideal solution has two components A and B. A is more volatile than B, i.e, PAo > PBo and also PAo > Ptotal . If
XA and YA are mole fractions of components A in liquid and vapour phases, then
(A) XA = YA
(B) XA > YA
(C) XA < YA
(D) Data insufficient
[Ans:- C]
7. Two liquids A and B have vapour pressure in the ratio PAo : PBo = 1: 3 at a certain temperature Assume A and B
form an ideal solution and the ratio of mole fractions of A to B in the vapour phase is 4 : 3. Then the mole
fraction of B in the solution at the same temperature is:
1
1
2
4
(A)
(B)
(C)
(D)
5
3
5
4
[Ans:- A]
CHAPTER PRACTICE PROBLEMS
CHEMISTRY
FIITJEE
8. Two liquids A and B have PAo and PBo in the ratio of 1 : 3 and the ratio of number of moles of A and B In liquid
phase are 1 :3 then mole fraction of ‘A’ in vapour phase in equilibrium with the solution is equal to:
(A) 0.1
(B) 0.2
(C) 0.5
(D) 1.0
[Ans:- A]
-1
9. Calculate the mass of the solute of molar mass 342 g mol that should be dissolved in 150g of H2O to reduce
0
its vapour pressure to 22.8 torr. The vapour pressure of pure water at 25 C is 23.75 torr
Ans: [114 g]
10. The vapour pressure of pure benzene at a certain temperature is 640 mm Hg. A non-volatile non-electrolyte
solid weighing 2.175 g is added to 39 g of benzene. The vapour pressure of the solution is 600 mm Hg. What
is the molecular weight of the solid substance?
[Ans: 65.25]
o
11. Calculate vapour pressure of a 5% (by weight) solution of water in glycerol (molecular weight 92.1) at 100 C.
assuming Raoult’s law to be valid and neglecting the vapour pressure of glycerol.
[Ans: 161.2 mm]
12. Two liquids A and B form ideal solutions. At 300 K, the vapour pressure of a solution containing 1 mole of A
and 3 moles of B is 550 mm of Hg. At the same temperature, if one more mole of B is added to this solution,
the vapour pressure of the solution increases by 10 mm of Hg. Determine the vapour pressure of A and B in
their pure states.
[Ans: PAo = 400 mm PBo = 600 mm]
13. At 310 K, the vapour pressure of an ideal solution containing 2 moles of A and 3 moles of B is 550 mm of Hg.
At the same temperature. If one mole of B is added to this solution, the vapour pressure of solution increases
by 10 mm of Hg. Calculate the V.P. of A and B in their pure state.
[Ans: A = 460, B =610 mm]
Comprehension
As we go higher above the sea level, the atmospheric pressure decreases. The exact decrease in pressure with
increase is given by the following expression called barometric formula.
⎛ P ⎞ Mgh
2.303 log ⎜ ⎟ =
⎝ P0 ⎠ RT
P0 = pressure at sea level, P is pressure at height ‘h’
–3
–1
M = Average molecular mass of air (= 28.8 × 10 kg mol )
–1
–1 –1
g = 9.8 ms , R = 8.314 J mol k , T is temperature in Kelvin.
Boiling point of the liquid is the temperature at which vapour pressure of liquid becomes equal to atmospheric
pressure. It means boiling point depends upon external pressure acting on the liquid. As the temperature
increases, vapour pressure increase.
14. The boiling point of water at higher altitude is
0
0
(A) 100 C
(B) < 100 C
0
(C) > 100 C
0
(D) 0 C
[Ans:- B]
15. The boiling point of water in pressure cooker is
0
0
(A) 100 C
(B) > 100 C
0
(C) < 100 C
0
(D) 25 C
[Ans:- B]
16. The relation between vapour pressure (V.P.) of a liquid and external pressure
1
Pexternal
(A)V.P. α Pexternal
(B) V.P.α
(C) V.P. is independent of external pressure
(D) V.P. = Pexternal
[Ans:- C]
CHAPTER PRACTICE PROBLEMS
FIITJEE
CHEMISTRY
Freeing Point Depression
17. Minimum freezing point will be for 1 molal solution of each compound, assuming complete ionization in each
case :
(A) ⎡⎣Fe (H2 O )6 ⎤⎦ Cl3
(B) ⎡⎣Fe (H2 O )5 Cl⎤⎦ Cl2 .H2O
(C) ⎡⎣Fe (H2O )4 Cl2 ⎤⎦ Cl.2H2O
(D) ⎡⎣Fe (H2 O )3 Cl3 ⎤⎦ .3H2 O
[Ans:- A]
18. Maximum freezing point will be for 1 molal solution of assuming equal ionisation in each case:
(A) ⎡⎣Fe (H2 O )6 ⎤⎦ Cl3
(B) ⎡⎣Fe (H2 O )5 Cl⎤⎦ Cl2 .H2O
(C) ⎡⎣Fe (H2O )4 Cl2 ⎤⎦ Cl.2H2O
(D) ⎡⎣Fe (H2 O )3 Cl3 ⎤⎦ .3H2 O
[Ans:- D]
19. Pt Cl4 .6H2O can exist as a hydrated complex. Its 1 molal aq. Solution has depression in freezing point of 3.72.
-1
Assume 100% ionization and k f (H2O) = 1.86o C mol kg, then complex is :
(A) ⎡⎣Pt (H2 O )6 ⎤⎦ Cl4
(B) ⎡⎣Pt (H2O )4 Cl2 ⎤⎦ Cl2 .2H2O
(C) ⎡⎣Pt (H2O )3 Cl3 ⎤⎦ Cl.3H2O
(D) ⎡⎣Pt (H2 O )2 Cl4 ⎤⎦ .4H2O
[Ans:- C]
20. A complex is represented as CoCl3 xNH3 its 0.1 molal solution in water shows ΔTf = 0.558 K. k f for H2 O is
-1
1.86 K molality . Assuming100% ionization of complex and co-ordination number of Co is six, calculate
formula of complex:
(A) ⎡⎣Co (NH3 )6 ⎤⎦ Cl3
(B) ⎡⎣Co (NH3 )5 Cl ⎤⎦ Cl2
(C) ⎡⎣Co (NH3 )4 Cl2 ⎤⎦ Cl
(D) None of these
21. The freezing point of equimolal aqueous solutions will be highest for:
(A) C6H5NH3Cl
(B) Ca (NO3 )2
(C) La (NO3 )2
[Ans:- B]
(D) C6H12O6
[Ans:- D]
22. The freezing point of a 4% aqueous solution of ‘A’ is equal to the freezing point of 10% aqueous solution of
‘B’. If the molecular weight of ‘A’ is 60, then molecular weight of ‘B’ will be:
(A) 160
(B) 90
(C) 45
(D) 180
[Ans:A]
23. The freezing point of aqueous solution that contains 5% by mass urea, 1.0% by mass KCI and 10% by mass
of glucose is [Kf (H2O) = 1.86 K Kg / mol]
(A) 290.2 K
(B) 285 .5 K
(C) 269.93 K
(D) 250 K
[Ans: C]
24. 20 g of a binary electrolyte (mol. Wt. = 100) are dissolved in 500 g of water. The freezing point of the solution
O
-1
is – 0.74 C. Kf = 1.86K kg mol . The degree of ionization of the electrolyte is
(A) 50%
(B) 75%
(C) 100%
(D) 0%
[Ans: D]
25. To 10 ml of 0.1 M BaCI2 solution, 0.2 ml of 3M Na2SO4 is added, BaSO4 is formed and is completely
insoluble. All the salts are completely dissociated and change in volume is negligible. The
(A) Freezing point is raised
(B) Freezing point is lowered
(C) Freezing point does not change
(D) boiling pint is lowered
[Ans: B]
CHAPTER PRACTICE PROBLEMS
FIITJEE
CHEMISTRY
26. The molarity and molality of an aqueous solution of H2SO4 as M1 and M2
respectively. If the molecular mass of H2SO4 is M, the density of the solution is
⎛ 1
M ⎞
+
⎟ g / ml
⎝ M1 1000 ⎠
M ⎞
⎛
(A) M2 ⎜ M1 +
g / ml
1000 ⎟⎠
⎝
(B) M2 ⎜
⎛ 1
M ⎞
+
⎟ g / ml
⎝ M2 1000 ⎠
M ⎞
⎛
(D) M1 ⎜ M2 +
g / ml
1000 ⎟⎠
⎝
(C) M1 ⎜
[Ans: B]
Boiling Point Elevation
27. Which one of the following aqueous solution will exhibit highest boiling point:
(A) 0.015 M urea
(B) 0.01 M KNO3
(C) 0.10 M Na2SO4
(D) 0.015 M glucose
[Ans:C]
28. Calculate the percentage degree of dissociation of an electrolyte XY2 (Normal molar mass =164) in water if
the observed molar mass by measuring elevation in boiling point is 65.6:
(A) 75%
(B) 25%
(C) 65%
(D) None of these
[Ans: A]
29. How much ethyl alcohol must be added to 1.0 L of water so that solution will not freeze
at −4o F? K f = 1.86o C / m
(
(A) < 20 g
)
(B) < 10.75 g
(C) <494.5 g
(D) >494.5 g
[Ans: D]
30. An aqueous solution of 2 per cent (w/w) non-volatile solute exerts a pressure of 1.004 bar at the boiling point
of the solvent. What is the molecular mass of the solute?
-1
[Ans: 40.97g mol ]
o
31. The latent heat of vaporization of water is 9700 cal/mol and if the boiling point of water is 100 C, the elevation
of boiling point in molal glucose solution is
(A) 0.516 K
(B) 1.026 K
(C) 10.26 K
(D) 1.832 K
[Ans: A]
32. An aqueous solution containing 5% by weight of urea and 10% by weight of glucose. What will be its freezing
-1
point? Kf for H2O is 1.86 mol kg.
o
[Ans: -3.04 C]
Osmosis
33. At 300K, two solutions of glucose in water of concentration 0.01 M and 0.001 M are separated by
semipermeable membrane with respect of water. On which solution, the pressure need be applied to prevent
osmosis? Calculate magnitude of this applied pressure.
[Ans: 0.01 M solution, 0.2217 atm pressure]
34. 0.85% aqueous solution of NaNO3 is apparently 90% dissociated. The osmotic pressure of solution at 300 K
is
(A) 4.674 atm
(B) 46.74 atm
(C) 2.46 atm
(D) 4.674 mm Hg
[Ans: D]
35. The solution containing 4.0 g of polyvinyl chloride polymer in 1 lit of dioxane was found to have an osmotic
-4
pressure 6.0 × 10 atm at 300 K. the mol. Mass of polymer was found to be
-2
5
(A) 3.0 × 10
(B) 1.6 × 10
4
2
(C) 5.6 × 10
(D) 6.4 × 10
[Ans: B]
CHAPTER PRACTICE PROBLEMS
FIITJEE
CHEMISTRY
Comprehension
Osmosis is the movement of solvent particles from a region of low concentration to a region of high concentration
or from a region of high vapour pressure to a region of low vapour pressure
A
H2O 0.1M
B
H2O 0.1M
X
Y
36. It is clear from the diagram that there is more osmosis in A in (X) than in B in (Y). Hence
(A) A = NaCl, B = KCl
(B) A = K4[Fe(CN)6], B = K3[Fe(CN)6]
(C) A = NaNO3, B = CuSO4
(D) A = AlCl3, B = FeCl3
[Ans:- B]
37. The degree of association related to molecular masses by the expression
(where M0 & Mn are the theoretical and experimentally observed molecular masses of solute respectively)
(A) α =
Mn − Mo
M0 (n − 1)
(B) α =
n (Mn − Mo )
M0 (1 − n )
(C) α =
M0 − Mn
Mn (n − 1)
(D) α =
n (Mn − M0 )
M0 (n − 1)
[Ans:- D]
38. The red blood cell swells in an aqueous solution this shows that the osmotic pressure of the solution inside
the blood cell is
(A)lower than the aqueous solution
(B) higher than the aqueous solution
(C)equal to aqueous solution
(D) none of these
[Ans:- B]
Vant Hoff Factor
39. The degree of dissociation of an electrolyte is α and its Van’t Hoff factor is i. The number of ions obtained by
complete dissociation of 1 molecule of the electrolyte is:
i + α −1
i −1
i + 1+ α
(A)
(B) i – α – 1
(C)
(D)
α
α
1− α
[Ans:- A]
40. One mole of a solute A is dissolved in a given volume of a solvent. The association of the solute take place
as follows:
nA É An
If α is the degree of association of A, the van’t Hoff factor i is expressed as:
α
1− α +
α
n
(A) i = 1 – α
(B) i = 1 +
(C) i =
n
1
(D) i = 1
[Ans:- C]
41. The degree of dissociation of Ca(NO3)2 in a dilute aqueous solution containing 7 g of the salt per 100gm of
o
o
water at 100 C is 70%. If the vapor pressure of water at 100 C is 760 mm. calculate the vapor pressure of
the solution.
[Ans: 746.10 mm]
42. The ratio of the value of any colligative property for KCI solution to that for sugar is nearly _________times.
(A) 1
(B) 0.5
(C) 2
(D) 2.5
[Ans: C]
43. The mol. Weight of NaCI determined by studying freezing point depression of its 0.5% aqueous solution is 30.
The apparent degree of dissociation of NaCI is
(A) 0.95
(B) 0.5
(C) 0.6
(D) 0.3
[Ans: A]
44. Match Column – I with Column – II
CHAPTER PRACTICE PROBLEMS
FIITJEE
Column – I( Aqueous solution)
CHEMISTRY
Column – II(van't Hoff factor)
(A) K4[Fe(CN)6]
(P) 1 + α
(B) Al2(SO4)3
(Q) Greater than 1
(C) NH2 CONH2
(D) CaCl2
(R) (1 + 4α)
(S) 1
[Ans : A : Q,R; B : Q,R; C : S; D : Q]
Ostwald Wacker Process
45. A current of dry air is passed through a solution of 5g of a in 100 gms of water and through water alone. The
loss in weight of solution and solvent are 0.78 g of 0.02 g. Calculate the molecular weight of A.
46. Dry air was passed through 5g of solute in 80 g of water and then through pure water. The loss in weight of
solution was 2.5 g and that of pure water was 0.04g. The molecular weight of solute is (in g/mol)
(A) 7.143
(B) 71.43
(C) 74.3
(D) none of these
Comprehension
Lowering of vapour pressure is determined by Ostwald and Walker dynamic method. It is based on the principle
that when air is allowed to pass through a solvent or solution, it takes up solvent vapour with it to get itself
saturated with vapour at that temperature
dry air
Solution(B)
PA
W1 loss
Solvent(A)
PAo
Anhydrous CaCl2
W2 loss
air
A and B are weighed separately before and after passing dry air. Loss in weight of each set, gives the lowering of
vapour pressure. The temperature of air, the solution and the solvent is kept constant.
47. Loss in weight of solvent (w2) will be proportional to
(A) PA − PAo
(B) PAo − PA
(C)
PA
PAo
(D) none of these
[Ans: B]
48.
PAo − PA
is equal to
PA
(A)
w1
w1 + w 2
(B)
w2
w1 + w 2
(C)
w1
w1 − w 2
(D)
w2
w1
[Ans: D]