Question

$1$ mole of ice at $0^{\circ }C$ and $4.6$ mm Hg pressure is converted to water vapour at a constant temperature and pressure. Find $\Delta H$ if the latent heat of fusion of ice is $80$ Cal/gm and latent heat of vaporization of liquid water at $0^{\circ }C$ is $596$ Cal per gram and the volume of ice in comparison to that of water (vapour) is neglected.

• A. 14.53 Kcal/mol
• B. 12.16 Kcal/mol
• C. 10.22 Kcal/mol
• D. 15.62 Kcal/mol

Answer 1

Answer: (B)

12.16 Kcal/mol

$\text{ice}⟶\text{vapour}$

Given that:-

$\Delta H_f=80cal/gm=80\times 18=1.44kcal/mol$

$\Delta H_v=596cal/gm=596\times 18=10.728Kcal/mol$

$\Delta H=\Delta H_f+\Delta H_v$

$\Delta H=1.44+10.728=12.168Kcal/mol$

Hence, the correct option is $\text{B}$