Standard enthalpy of vapourisation for $\mathrm{CCl}_4$ is $30.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$. Heat required for vapourisation of $284 \mathrm{~g}$ of $\mathrm{CCl}_4$ at constant temperature is ________ $\mathrm{kJ}$.
(Given molar mass in $\mathrm{g} \mathrm{mol}^{-1} ; \mathrm{C}=12, \mathrm{Cl}=35.5$)
Answer (integer)
56
Solution
<p>$$\begin{aligned}
& \Delta \mathrm{H}_{\text {vap }}^0 \mathrm{CCl}_4=30.5 \mathrm{~kJ} / \mathrm{mol} \\
& \text { Mass of } \mathrm{CCl}_4=284 \mathrm{~gm} \\
& \text { Molar mass of } \mathrm{CCl}_4=154 \mathrm{~g} / \mathrm{mol} \\
& \text { Moles of } \mathrm{CCl}_4=\frac{284}{154}=1.844 \mathrm{~mol} \\
& \Delta \mathrm{H}_{\text {vap }}{ }^{\circ} \text { for } 1 \mathrm{~mole}=30.5 \mathrm{~kJ} / \mathrm{mol} \\
& \Delta \mathrm{H}_{\text {vap }}{ }^{\circ} \text { for } 1.844 \mathrm{~mol}=30.5 \times 1.844 \\
& \quad=56.242 \mathrm{~kJ}
\end{aligned}$$</p>
About this question
Subject: Chemistry · Chapter: Thermodynamics · Topic: Zeroth and First Law
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