The vapour pressure of pure benzene at 25° C is 640 mm Hg and that of the solute A in benzene is 630 mm of Hg. The molality of solution is

Question

The vapour pressure of pure benzene at 25° C is 640 mm Hg and that of the solute A in benzene is 630 mm of Hg. The molality of solution is (A) 0.2 m (B) 0.4 m (C) 0.5 m (D) 0.1 m

Tardigrade Admin · Accepted Answer

textP text= textP texto textX textA 630=640XA XA=(630/640)=0.984 XB=0.016 m=(XB × 1000/XA MA)=(0.016 × 1000/0.984 × 78) =0.2 m

Q. The vapour pressure of pure benzene at $2 5^{\circ} C$ is 640 mm Hg and that of the solute A in benzene is 630 mm of Hg. The molality of solution is

0.2 m

0.4 m

0.5 m

0.1 m

$P = P^{o} X_{A}$

$630 = 640 X_{A}$

$X_{A} = \frac{630}{640} = 0.984$

$X_{B} = 0.016$

$m = \frac{X _{B} \times 1000}{X _{A} M _{A}} = \frac{0.016 \times 1000}{0.984 \times 78}$

$= 0.2 m$

Q. The vapour pressure of pure benzene at 25∘C is 640 mm Hg and that of the solute A in benzene is 630 mm of Hg. The molality of solution is