Liquid Solution

Q: The vapour pressures of ethanol and methanol are 44.5mm and 88.7mm Hg respectively at the same temperature. An ideal solution is formed by mixing 60gm of ethanol and 40gm of methanol. The mole fraction of methanol in the vapour phase is.

(A) 0.66

(B) 0.55

(C) 0.11

(D) 0.33

Solution: Mole of CH₃CH = 40/32 = 1.25

Mole of CH₃CH₂OH = 60/46 = 1.304

Let CH₃CH → A, C₂H₅OH → B

$\large X_A = \frac{n_A}{n_A + n_B} = \frac{1.25}{1.25 + 1.304} $ XA

= 0.49

∴ X_B = 0.51

$\large P_T = P^o_A X_A + P^o_B X_B $

= 88.7 × 0.49 + 44.5 × 0.51

P_T = 43.480 + 22.695 = 66.175

P_A = 43.48

X_A = P_A /P_T = 43.48/66.175

= 0.66

Answer is (A)

Q: If the total vapour pressure of the liquid mixture A and B is given by the equation: P = 180 X_A + 90 then the ratio of the vapour pressure of the pure liquids A and B is given by:

(A) 3:2

(B) 4:1

(C) 3:1

(D) 6:1

Solution: As we know

$\large P_T = P^o_A X_S + P^o_B X_A $

$\large = P^o_A X_A + P^o_B (1- X_A ) $

$\large = (P^o_A – P^o_B )X_A + P^o_B $ …(i)

But from question,

P_T = 180 X_A + 90 ….(ii)

Equating equation (i) and (ii)

$\large (P^o_A – P^o_B ) = 180 $

P^o_B = 90 (when xA = 0 and xB = 1, P = P^o_B or P^o_B = 0 + 90 = 90)

P^o_A = 180 + 90 = 270

P^o_A / P^o_B =270/90 = 3:1

Answer is (C)

Q: NaCl is added to 1 litre water to such an extent that ΔT_f/K_f becomes to 1/500 , the wt. of NaCl added is

(A) 5.85 gm

(B) 0.585 gm

(C) 0.0585 gm

(D) None of these

Solution: For NaCl , i =2

∴ ΔT_f = K_f × i × C_m

ΔT_f/K_f = 2 × Cm

or , 1/500 = 2 × molality

or molality = 1/1000

i.e., 1/1000 moles Kg^–1 water

i.e., 1/1000 moles per litre water

= 1/1000 × 58.5 gm NaCl

= 0.0585 gm NaCl

Answer is (C)

Q: The vapour pressure of solvent is 20 torr, while that of its dilute solution is 17 torr, the mole-fraction of the solvent is

(A) 0.6

(B) 0.85

(C) 0.5

(D) 0.7

Solution: From question

P° = 20 torr, P_S = 17 torr

$\large \frac{P^o – P_S}{P^o} = X_{solute}$

3/20 = X_solute

or , X_{solute = 0.15}

X_solvent = 1 – 0.15 = 0.85

Q: Which one of the following pairs of solution will be isotonic at the same temperature ?

(A) 1 M NaCl & 2 M – Urea

(B) 1 M CaCl₂ & 1.5 M – KCl

(C) 1.5 M AlCl₃ & 2 M Na₂SO₄

(D) 2.5 M KCl and 1 M – Al₂(SO₄)₃

Solution: For
(A) 1M NaCl solution (i × C_m) = 2 × 1M (C_m = Molarity)

2M Urea solution (i × C_m) = 2M

1M CaCl₂ solution (i × C_m) = 3 × 1M = 3M

1.5 M KCl solution (i × C_m) = 2 × 1.5M = 3M

Similarly for 1.5 M AlCl₃ and 2M Na₂SO₄

The value of i × C_m be 6

and 2.5 M KCl and 1M Al₂(SO₄)₃ has 5 M each.

So, the each pair of solution

A,B,C and D have same concentration so, all pairs of solution be isotonic,

Answer : A,B,C and D

Q: If P₀ & P_S be the vapour pressure of solvent and its solution respectively and N₁ and N₂ be the mole-fractions of solvent and solute respectively , then:

(A) P_S = P₀ N₂

(B) P₀ = P_S = P . N₂

(C) P_S = P₀ N₁

(D) $\large \frac{P_0 – P_S}{P_S} = \frac{N_1}{N_1 + N_2} $

Solution: According to Raoult’s Laws

Vapour pressure of solution = V.P. of solvent X solvent

$\large \frac{P_0 – P_S}{P_0} = N_2 $

$\large P_0 – P_S = P_0 \times N_2 $

$\large P_S = P_0 – P_0 \times N_2 $

$\large P_S = P_0 (1 – N_2 ) = P_0 N_1 $ (Since , N₁ + N₂ = 1)

Answer is (C)

Q: For a binary ideal liquid solution, the total pressure of the solution is given as

(A) $\large P_{total} = P^∗_A + (P^∗_A – P^∗_B)X_A $

(B) $\large P_{total} = P^∗_B + (P^∗_A – P^∗_B)X_A $

(C) $\large P_{total} = P^∗_A + (P^∗_B – P^∗_A)X_A $

(D) $\large P_{total} = P^∗_B + (P^∗_A – P^∗_B)X_A $

Solution : Since we know that

$\large P_T = P^∗_A X_A + P^∗_B X_B $

$\large = P^∗_A X_A + P^∗_B ( 1- X_A ) $

$\large P_{total} = P^∗_B + (P^∗_A – P^∗_B ) X_A $

Answer is (B)