Molar heat of vaporization of a liquid is $6\mathrm{kJ}{\mathrm{mol}}^{-1}$ if the entropy change is$16{\mathrm{Jmol}}^{-1}{\mathrm{K}}^{-1}$, the boiling point of the  liquid is 

Since boiling occurs at equilibrium conditions, so ΔG=0

Now, T =$\frac{∆\mathrm{H}}{∆\mathrm{S}}$​
T = $\frac{6000\mathrm{J}/\mathrm{mol}}{16 \mathrm{J}/\mathrm{mol}/\mathrm{K}}$
T = 375K
Hence, the boiling point of the liquid is 375 K.