Question

the kb and kf values of a liquid is 0.6 and 1.8 respectively. its boiling point and freezing point is 300k and 200k respectively. its enthalpy of vapourization is x times the enthalpy of fusion .then the value of x is?

Answer 1

Given:

The value of $K_b$ of the liquid = 0.6

The value of $K_f$ of the liquid = 1.8

The boiling point of the liquid = 300 K

The freezing point of the liquid = 200 K

The enthalpy of vapourization is x times the enthalpy of fusion .

To Find :

The value of x = ?

Solution :

∴ We know that $K_f = \frac {R \times M_1 \times T_f^2}{1000 \times \Delta H_{fus}}$

And also $K_b = \frac{R \times M_1 \times T_b^2}{1000 \times \Delta H_{vap}}$

Here , R is universal gas constant , $M_1$ is molar mass of the element , $T_f$ is freezing point , $\Delta H _{fus}$ is enthalpy of fusion , $T_b$ is boiling point and $\Delta H_{vap}$ is enthalpy of vapourisation .

So , $\frac{K_f}{K_b}=\frac{(\frac{R\times M_1 \times T_f^2}{1000 \times \Delta H _{fus}})}{(\frac{R\times M_1 \times T_b^2}{1000 \times \Delta H _{vap}})}$

Or, $\frac{1.8}{0.6} =(\frac{T_f}{T_b})^2\times \frac{\Delta H_{vap}}{\Delta H_{fus}}$

Or, 3 = $(\frac{200}{300})^2 \times \frac{x. \Delta H_{fus}}{\Delta H _{fus}}$

Or,  3 = $\frac{4}{9}x$

Or, $x = \frac{27}{4}$

So, x = 6.75

Hence , the value of x is 6.75