The enthalpy of vapourisation of $CCl_4~ 30.5~kJ~mol^{-1}$ is Calculate the heat required for the vapourisation of $284~g$ of $CCl_4$ at constant pressure. (Molar mass of $CCl_4 = 154~g~mol^{-1}$)

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Solution

The enthalpy of vapourisation $(\Delta H)$ of $CCl_4 = 30.5~kJ~mol^{-1}$

$q_p = \Delta H = 30.5~kJ~mol^{-1}$

No.of moles of $CCl_4 = \dfrac{given~mass(g)}{molar~mass~(g~mol^{-1})}$

$=\left ( \dfrac{284~g}{154~g~mol^{-1}} \right )$

$= 1.844~mol$

Hence, the heat required for the vapourisation of $1.844~mol$ of $CCl_4$ at constant pressure

$= 1.844~mol \times 30.5~kJ~mol^{-1}$

Heat required $= 56.24~kJ$

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