Chemistry - Chemical Equilibrium Question with Solution | TestHub

The degree of association, $\alpha =\frac{2\left({\mathrm{K}}_{\mathrm{b}}\mathrm{t}-\Delta {\mathrm{T}}_{\mathrm{b}(\text{ obs })}\right)}{{\mathrm{K}}_{\mathrm{b}}\mathrm{t}}$

The molecular mass observed will be less than actual molecular mass.

Colligative properties observed will be $\Delta {\mathrm{T}}_{\mathrm{b}}(\mathrm{obs})<\Delta {\mathrm{T}}_{\mathrm{b}}$ (theoretical);

$\Delta {\mathrm{T}}_f(\mathrm{obs})<\Delta {\mathrm{T}}_f$ (theoretical)

The equilibrium constant for the process can be expressed as

$$K_{eq}=\frac{{\mathrm{K}}_{\mathrm{b}}\mathrm{t}\left({\mathrm{K}}_{\mathrm{b}}\mathrm{t}-\Delta {\mathrm{T}}_{\mathrm{b}}\right)}{[2\Delta {\mathrm{T}}_{\mathrm{b}}-{\mathrm{K}}_{\mathrm{b}}\mathrm{t}{]}^2}$$