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).
Show Hint
The boiling point elevation constant depends on both the square of the boiling point and the inverse of the enthalpy of vaporization.
Correct Answer: 8
Solution and Explanation
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.
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