Question

The molar entropy of vaporization of acetic acid is; 14.4 cal $K^{-1}$ $mo{l}^{-1}$ at its boiling point ${18}^{\circ }C$. The latent heat of vaporisation of acetic acid is:

• A. 49 cal $g^{-1}$
• B. 69.84 cal $g^{-1}$
• C. 94 cal $g^{-1}$
• D. 84 cal $g^{-1}$

Answer 1

The correct option is B 69.84 cal $g^{-1}$

From second law of thermodynamics,

$ds=\frac{dq}{T}$ or $\Delta S=\frac{\Delta q}{T}$

For vapourisation,

$ds⟶$ molar entropy of vpourisation

$dq⟶$ latent heat of vapourisation

$T⟶$ boiling point

$dq=Tds=291k\times 14.4cal{K}^{-1}mo{l}^{-1}=4190.4calmo{l}^{-1}$

For $1mol$ of $C{H}_{3}COOH$, $\Delta q=4190.4cal$

For $60g$ of $C{H}_{3}COOH$, $\Delta q=4190.4cal$

For $1g$ of $C{H}_{3}COOH$, $\Delta q=\frac{4190.4}{60}cal{g}^{-1}=69.84cal{g}^{-1}$