Question

The enthalpy of vaporisation of a substances is 840 J $mol^{- 1}$ and its boiling point is $-173^\circ C$ . Its enthalpy of vaporisation is

• A. $8.4$ $Jmol^{- 1}K^{- 1}$
• B. $49$ $mol^{- 1}K^{- 1}$
• C. $21$ $mol^{- 1}K^{- 1}$
• D. $12$ $mol^{- 1}K^{- 1}$

Answer 1

Answer: (A)

$\Delta S_{v a p}=\frac{q_{r e v}}{T}=\frac{\Delta H_{v a p}}{T}$

$=\frac{840}{\left(\right. - 173 + 273 \left.\right) K}$

$=8.4Jmol^{- 1}K^{- 1}$ .