Liquids A and B form an ideal solution and the former has stronger intermolecular forces. If xA and x′A are the mole fractions of A in the solution and vapour in equilibrium, then

Question

Liquids A and B form an ideal solution and the former has stronger intermolecular forces. If xA and x′A are the mole fractions of A in the solution and vapour in equilibrium, then (A)  (x'A/xA) = 1  (B)  (x'A/xA) > 1  (C)  (x'A/xA) < 1  (D)  x'A + xA = 1

Tardigrade Admin · Accepted Answer

x'A = (pA/p) = (p°AxA/p) ⇒ (x'/xA) = (p°A/p)° < 1 (since liquid A is less volatile, p°A < p)

Q. Liquids A and B form an ideal solution and the former has stronger intermolecular forces. If $x_{A}$ and $x'_{A}$ are the mole fractions of A in the solution and vapour in equilibrium, then

$\frac{x _{A}^{'}}{x _{A}} = 1$

$\frac{x _{A}^{'}}{x _{A}} > 1$

$\frac{x _{A}^{'}}{x _{A}} < 1$

$x_{A}^{'} + x_{A} = 1$

$x_{A}^{'} = \frac{p _{A}}{p} = \frac{p _{A}^{\circ} x _{A}}{p} \Rightarrow \frac{x ^{'}}{x _{A}} = \frac{p _{A}^{\circ}}{p}^{\circ} < 1$
(since liquid A is less volatile, $p_{A}^{\circ} < p$ )

Q. Liquids A and B form an ideal solution and the former has stronger intermolecular forces. If xA​ and x′A​ are the mole fractions of A in the solution and vapour in equilibrium, then

xA​xA′​​=1

xA​xA′​​>1

xA​xA′​​<1

xA′​+xA​=1