If the boiling points of two solvents X and Y | vedclass

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(A) The boiling point elevation constant $K_B$ is given by the formula: $K_B = \frac{R \cdot T_B^2 \cdot M}{1000 \cdot \Delta H_{vap}}$.Since the mole

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(A) The boiling point elevation constant $K_B$ is given by the formula: $K_B = \frac{R \cdot T_B^2 \cdot M}{1000 \cdot \Delta H_{vap}}$.Since the molecular weights $M$ are the same and $R$ is a constant,we have $K_B \propto \frac{T_B^2}{\Delta H_{vap}}$.Given $\frac{(T_B)_X}{(T_B)_Y} = \frac{2}{1}$ and $\frac{(\Delta H)_X}{(\Delta H)_Y} = \frac{1}{2}$.Therefore,$\frac{(K_B)_X}{(K_B)_Y} = \left( \frac{(T_B)_X}{(T_B)_Y} ight)^2 	imes \frac{(\Delta H)_Y}{(\Delta H)_X}$.Substituting the values: $\frac{(K_B)_X}{(K_B)_Y} = (2)^2 	imes \left( \frac{2}{1} ight) = 4 	imes 2 = 8$.Thus,$m = 8$.

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