Question

Molar heat of vaporisation of a liquid is $6 \,kJ\, mol^{-1}$. If the entropy change is $16 \,J \,mol^{-1}\, K^{-1}$, the boiling point of the liquid is

• A. $375^?C$
• B. $375\,K$
• C. $273\,K$
• D. $102^?C$

Answer 1

Answer: (B)

$375\,K$

Answer 2

$\Delta S=16\,J \, mol^{-1}\, K^{-1}$ $T_{b.p.}=\frac{\Delta H_{vap}}{\Delta S_{vap}}=\frac{6\times1000}{16}=375\,K$