## 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

$\large P_T = P^∗_A X_A + P^∗_B X_B$

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

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

## Which of the following plots represents the behaviour of an ideal binary liquid solution ?

Q: Which of the following plots represents the behaviour of an ideal binary liquid solution ?

(A) Plot of Ptotal vs YA (mol-fraction of A in vapour phase) is linear

(B) Plot of Ptotal vs YB is linear

(C) Plot of 1/Ptotal vs YA is linear

(D) Plot of 1/Ptotal vs YB is non-linear

Solution: $\large Y_A = \frac{P^0_A X_A}{P_{total}}$

Graph of YA Vs $\frac{1}{P_{total}}$is linear

## The vapour pressure of benzene at 80°C is lowered by 10 mm by dissolving 2g of a non-volatile substance in 78g of benzene…

Q: The vapour pressure of benzene at 80°C is lowered by 10 mm by dissolving 2g of a non-volatile substance in 78g of benzene. The vapour pressure of pure benzene at 80°C is 750 mm. The molecular weight of the substance will be:

(A) 15

(B) 150

(C) 1500

(D) 148

Solution: $\large \frac{P_0 – P_S}{P_0} = \frac{w \times M}{m \times W}$

$\large \frac{10}{750} = \frac{2 \times 78}{m \times 78}$

m = 150

## Solute A is a ternary electrolyte and solute B is non-electrolyte. If 0.1 M solution of solute B produces an osmotic pressure of 2P …

Q: Solute A is a ternary electrolyte and solute B is non-electrolyte. If 0.1 M solution of solute B produces an osmotic pressure of 2P, then 0.05 M solution of A at the same temperature will produce on osmotic pressure equal to

(A) P

(B) 1.5 P

(C) 2P

(D) 3P

Solution: For ternary electrolyte;

P1 = CST = 0.05 × 3 × S × T

For B; 2P = 0.1 × S × T;

P1 = 3P