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

Question

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

Tardigrade Admin · Accepted Answer

(P0-Ps/Ps)=(w× M/m× W) (10/(750-10))=(2×78/m×78) ∴ m=148 ; NOTE: [m comes 150 if formula (P0-Ps/P0)=(w × M/m× W) is used. But this is only for dilute solutions]

Q. The vapour pressure of benzene at $80° C$ is lowered by $10 mm$ by dissolving $2 g$ of a non-volatile substance in 78 g of benzene. The vapour pressure of pure benzene at $8 0^{\circ} C$ is $750 mm$ . The molecular weight of the substance will be:

$15$

$150$

$1500$

$148$

$\frac{P _{0} - P _{s}}{P _{s}} = \frac{w \times M}{m \times W}$
$\frac{10}{( 750 - 10 )} = \frac{2 \times 78}{m \times 78}$
$∴ m = 148$ ;
NOTE: [m comes 150 if formula $\frac{P _{0} - P _{s}}{P _{0}} = \frac{w \times M}{m \times W}$ is used. But this is only for dilute solutions]

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

15

150

1500

148