Question

The vapour pressure of pure benzene at ${50}^{\circ }C$ is $268torr$. How many mols of non-volatile solute per mol of benzene is required to prepare a solution of benzene having a vapour pressure of $167torr$at${50}^{\circ }C$:

• A. 0.377
• B. 0.605
• C. 0.623
• D. 0.395

Answer 1

The correct option is A 0.377

Vapour Pressure of pure benzene is 268 torr

Required Vapour pressure = 167 torr

According to Raoult’s law

$\frac{\Delta P}{P\circ }=\frac{\text{Moles of solute}}{\text{Moles of Solvent}}$

$\frac{\text{Moles of solute (Nb)}}{\text{Moles of Solvent (Na)}}=\frac{268–167}{268}$

$\frac{\text{Moles of solute (Nb)}}{\text{Moles of Solvent (Na)}}=0.3771$

Hence 1 mole of Benzene contains = 0.377 moles of solute.

Hence, the correct option is $A$