## The vapour pressures of ethanol and methanol are 44.5mm and 88.7mm Hg respectively at the same temperature…

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

∴ XB = 0.51

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

= 88.7 × 0.49 + 44.5 × 0.51

PT = 43.480 + 22.695 = 66.175

PA = 43.48

XA = PA /PT = 43.48/66.175

= 0.66

## If the total vapour pressure of the liquid mixture A and B is given by the equation: P = 180 XA + 90 then the ratio of…

Q: If the total vapour pressure of the liquid mixture A and B is given by the equation: P = 180 XA + 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,

PT = 180 XA + 90 ….(ii)

Equating equation (i) and (ii)

$\large (P^o_A – P^o_B ) = 180$

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

PoA = 180 + 90 = 270

PoA / PoB =270/90 = 3:1

## NaCl is added to 1 litre water to such an extent that ΔTf/Kf becomes to 1/500 , the wt. of NaCl added is

Q: NaCl is added to 1 litre water to such an extent that ΔTf/Kf 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

∴ ΔTf = Kf × i × Cm

ΔTf/Kf = 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

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

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, PS = 17 torr

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

3/20 = Xsolute

or , Xsolute = 0.15

Xsolvent = 1 – 0.15 = 0.85

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

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 × Cm) = 2 × 1M (Cm = Molarity)

2M Urea solution (i × Cm) = 2M

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

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

Similarly for 1.5 M AlCl3 and 2M Na2SO4

The value of i × Cm 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

## If P0 & PS be the vapour pressure of solvent and its solution respectively and N1 and N2 be the mole-fractions of solvent and solute respectively

Q: If P0 & PS 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) PS = P0 N2

(B) P0 = PS = P . N2

(C) PS = 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)

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

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

$\large P_T = P^∗_A X_A + P^∗_B X_B$

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

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