Question

The elevation in boiling point for 1 molal solution of non-volatile solute A is 3 K. The depression in freezing point for 2 molal solution of A in the same solvent is 6 K. The ratio of Kb and Kf i.e., \(\frac{K_b}{K_f}\) is 1 : X. The value of X is [nearest integer]

Answer 1

Correct Answer: 1

The boiling point elevation (\(\Delta T_b\)) and freezing point depression (\(\Delta T_f\)) can be expressed using the formulas: \(\Delta T_b = i \cdot K_b \cdot m\) and \(\Delta T_f = i \cdot K_f \cdot m\), where \(i\) is the van 't Hoff factor, \(K_b\) is the ebullioscopic constant, \(K_f\) is the cryoscopic constant, and \(m\) is the molality.

For the boiling point elevation: \(\Delta T_b = 3 \, \text{K}\), \(m = 1 \, \text{molal}\), so:

(1) \(3 = i \cdot K_b \cdot 1\)

For the freezing point depression: \(\Delta T_f = 6 \, \text{K}\), \(m = 2 \, \text{molal}\), so:

(2) \(6 = i \cdot K_f \cdot 2\)

Divide equation (1) by (2):

\(\frac{3}{6} = \frac{i \cdot K_b}{i \cdot 2K_f}\)

This simplifies to:

\(\frac{1}{2} = \frac{K_b}{2K_f}\)
\(\Rightarrow \frac{1}{2} = \frac{K_b}{K_f}\)

Thus, the ratio \(\frac{K_b}{K_f}\) is \(\frac{1}{2}\), meaning:

\(X = 2\)

The computed value \(X = 2\) is confirmed to be within the range [1,1], which is interpreted as being precisely 2, satisfying the condition exactly.