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 mass 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 mass of the substance will be : (A) 15 (B) 14.8 (C) 1500 (D) 148

Tardigrade Admin · Accepted Answer

( P °- P s / P °)=X text Benzene ⇒ (750-740/750)=(2 / M /(2) M +(78)78 On Solving, M=148

Q. The vapour pressure of benzene at $8 0^{\circ} C$ 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 mass of the substance will be :

15

14.8

1500

148

$\frac{P ^{\circ} - P _{s}}{P ^{\circ}} = X_{Benzene}$
$\Rightarrow \frac{750 - 740}{750} = \frac{2/ M}{\frac{2}{M} + \frac{78}{78}}$
On Solving, $M = 148$

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

15

14.8

1500

148

P∘P∘−Ps​​=XBenzene ​ ⇒750750−740​=M2​+7878​2/M​ On Solving, M=148

Q. The vapour pressure of benzene at $8 0^{\circ} C$ 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 mass of the substance will be :

$\frac{P ^{\circ} - P _{s}}{P ^{\circ}} = X_{Benzene}$
$\Rightarrow \frac{750 - 740}{750} = \frac{2/ M}{\frac{2}{M} + \frac{78}{78}}$
On Solving, $M = 148$