The enthalpy of vapourisation of a liquid is $30\mathrm{KJ}{\mathrm{mol}}^{-1}$ and entropy of vaporization is $75\mathrm{J}{\mathrm{mol}}^{-1}{\mathrm{k}}^{-1}$ . The boiling point of the liquid at 1atm pressure is.

 As the enthalpy given is $30\mathrm{KJ}{\mathrm{mol}}^{-1}$:

$\mathrm{\Delta H}=30\mathrm{KJ}{\mathrm{mol}}^{-1}=3000\text{ J/mol}$

At boiling point, the reaction will be at equilibrium, so:

$\mathrm{\Delta G}=0$ 

Now putting the values, we get:

$$\begin{gathered} \mathrm{\Delta G}=\mathrm{\Delta H}-\mathrm{T\Delta S} \\ \mathrm{T}=\frac{\mathrm{\Delta H}}{\mathrm{\Delta S}}=\frac{3000}{75}=400\text{ K} \end{gathered}$$

Therefore, the correct option is B.