Question

If the boiling points of two solvents X and Y (having same molecular weights) are in the ratio 2:1 and their enthalpy of vaporizations are in the ratio 1:2, then the boiling point elevation constant of X is m times the boiling point elevation constant of Y. The value of m is ______ (nearest integer).

Hint:

The boiling point elevation constant depends on both the square of the boiling point and the inverse of the enthalpy of vaporization.

Answer 1

Correct Answer: 8

The boiling point elevation constant is given by: 

\( K_b = \frac{RT_b^2 M}{1000 \Delta H_{\text{vap}}} \)

Where:

• \(R\): Universal gas constant
• \(T_b\): Boiling point
• \(M\): Molecular weight
• \(\Delta H_{\text{vap}}\): Enthalpy of vaporization

Step 1: Ratio of Boiling Point Elevation Constants

For solvents X and Y, taking the ratio of their boiling point elevation constants:

\( \frac{(K_b)_X}{(K_b)_Y} = \frac{\left(\frac{RT_b^2 M}{\Delta H_{\text{vap}}}\right)_X}{\left(\frac{RT_b^2 M}{\Delta H_{\text{vap}}}\right)_Y} \)

Step 2: Simplify

The equation simplifies to:

\( \frac{(K_b)_X}{(K_b)_Y} = \frac{(T_b^2)_X}{(T_b^2)_Y} \times \frac{(\Delta H_{\text{vap}})_Y}{(\Delta H_{\text{vap}})_X} \)

Step 3: Substitute Given Ratios

• \( \frac{(T_b)_X}{(T_b)_Y} = 2 \)
• \( \frac{(\Delta H_{\text{vap}})_X}{(\Delta H_{\text{vap}})_Y} = \frac{1}{2} \)

Substituting these values:

\( \frac{(K_b)_X}{(K_b)_Y} = \frac{2^2}{1^2} \times \frac{2}{1} \)

Step 4: Calculate

\( \frac{(K_b)_X}{(K_b)_Y} = \frac{4}{1} \times \frac{2}{1} = \frac{8}{1} \)

Conclusion:

The ratio of boiling point elevation constants, \( m \), is 8.