Question

The molar entropy of vaporisation of acetic acid is $14.4cal{K}^{-1}mo{l}^{-1}$ at its boiling point ${118}^{\circ }C$. The latent heat of vaporisation of acetic acid is:

• A. $49cal{g}^{-1}$
• B. $64cal{g}^{-1}$
• C. $94cal{g}^{-1}$
• D. $84cal{g}^{-1}$

Answer 1

Answer: (C)

$94cal{g}^{-1}$

Given, $E_V=14.4cal{K}^{-1}{mol}^{-1}$ $T=391K$

$1$ mol of ${CH}_{3}COOH=60g$

Hence, $L_V=E_{V}T$

$L_V=\frac{14.4}{60}\times 391\simeq 93.84cal{g}^{-1}=94cal{g}^{-1}$