Question

A substance 'X' (1.5 g) dissolved in 150 g of a solvent 'Y' (molar mass = 300 g mol$^{-1}$) led to an elevation of the boiling point by 0.5 K. The relative lowering in the vapour pressure of the solvent 'Y' is $____________ \(\times 10^{-2}\). (nearest integer) 
[Given : $K_{b}$ of the solvent = 5.0 K kg mol$^{-1}$] 
Assume the solution to be dilute and no association or dissociation of X takes place in solution.

Hint:

For dilute aqueous solutions, you can use the relation $RLVP = \frac{m \cdot M_{solvent}}{1000}$. This allows you to bypass calculating individual mole counts.

Answer 1

Correct Answer: 3

To find the relative lowering in the vapor pressure of the solvent 'Y', we begin by utilizing the colligative property of boiling point elevation. The formula for boiling point elevation is given by: \(\Delta T_b = i \cdot K_b \cdot m\), where \(i\) is the van't Hoff factor (1 for non-associating solute), \(K_b\) is the ebullioscopic constant of the solvent, and \(m\) is the molality of the solution. Given \(\Delta T_b = 0.5\) K and \(K_b = 5.0\) K kg mol\(^{-1}\), we calculate the molality: 

\(m = \frac{\Delta T_b}{K_b} = \frac{0.5}{5.0} = 0.1\) mol/kg.

Next, we calculate the moles of solute 'X'. Given that the mass of 'X' is 1.5 g, the molality formula \((m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}})\) allows us to determine the moles of 'X':

\(\text{moles of 'X'} = 0.1 \times 0.15 = 0.015\) mol.

Using the formula for relative lowering of vapor pressure \(\frac{\Delta P}{P^0} = \frac{\text{moles of solute}}{\text{moles of solvent}}\), we calculate the moles of solvent 'Y'. Given the mass of the solvent is 150 g \((0.15\) kg\() and its molar mass is 300 g mol\(^{-1}\)), we find:

\(\text{moles of 'Y'} = \frac{150}{300} = 0.5\) mol.

Thus,

\(\frac{\Delta P}{P^0} = \frac{0.015}{0.5} = 0.03\),

which translates to the relative lowering in vapor pressure as \(3 \times 10^{-2}\).

The calculated value \(3\) lies within the range (3,3), confirming correctness.