The vapour pressure of pure benzene is 639.7 mm Hg and the vapour pressure of a solution of a solute in benzene at the temperature is 631.9 mm Hg. If the molality of the solution is $156 \times 10^{- x}$ , then what is the value of $x$ ?

Open in App

Solution

Given,
Vapour pressure of pure benzene $(P^{o})$ =639.7 mm Hg

Vapour pressure of solution (P) =631.9 mm Hg

According to Raoult's law

$\frac{P^{o} - P}{P^{o}} = \frac{W_{2} \times M w_{1}}{M w_{2} \times W_{1}}$

where, $W_{2}$ and $W_{1}$ =weight of solute and solvent .respectively.

$M w_{2}, M w_{1}$ =Molecular weight of solute and solvent,respectively.

$∴ M o l a l i t y = \frac{W_{2}}{M w_{2} \times W_{1}} \times 1000$

$= \frac{P^{o} - P}{P^{o}} \times \frac{1}{M w} \times 1000$

$= \frac{639.7 - 631.9}{639.7} \times \frac{1}{78} \times 1000$

$= \frac{7.8 \times 1000}{639.7 \times 78} \times 1000 = 0.156 m o l {k g}^{- 1}$

Suggest Corrections

Similar questions

The vapour pressure of Benzene at $25^{\circ}$ is 639.7mm of Hg. The vapour pressure of a solution of benzene at same temperature is 631.9 mm of Hg. The molality of solution is:

Q. The V.P. of pure benzene at

25^{\circ} C

639.7 m m H g

and the V.P. of a solution of a non-volatile solute in benzene at the same temperature is

631.9 m m H g

. Calculate the mole fraction of solute and molality of the solution.

Q. The vapour pressure of pure benzene at a certain temperature is

640

mm Hg. A non-volatile solute weighing

2.175

g is added to

39.0

g of benzene. The vapour pressure of solution is

600

mm Hg. What is the molar mass of the solute?

Q. The vapour pressure of pure benzene at a certain temperature is

640 m m H g

. A non-volatile solid weighing

2.715 g

is added to

39.0 g

of benzene.

The vapour pressure of the solution is

600 m m H g

. What is the molar mass of the solid substance?

Join BYJU'S Learning Program

The vapour pressure of pure benzene is 639.7 mm Hg and the vapour pressure of a solution of a solute in benzene at the temperature is 631.9 mm Hg. If the molality of the solution is 156×10−x, then what is the value of x?

The vapour pressure of pure benzene is 639.7 mm Hg and the vapour pressure of a solution of a solute in benzene at the temperature is 631.9 mm Hg. If the molality of the solution is $156 \times 10^{- x}$ , then what is the value of $x$ ?