JEE MAIN - Chemistry (2024 - 29th January Evening Shift - No. 22)
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
56
Explanation
$$\begin{aligned}
& \Delta \mathrm{H}_{\text {vap }}^0 \mathrm{CCl}_4=30.5 \mathrm{~kJ} / \mathrm{mol} \\
& \text { Mass of } \mathrm{CCl}_4=284 \mathrm{~gm} \\
& \text { Molar mass of } \mathrm{CCl}_4=154 \mathrm{~g} / \mathrm{mol} \\
& \text { Moles of } \mathrm{CCl}_4=\frac{284}{154}=1.844 \mathrm{~mol} \\
& \Delta \mathrm{H}_{\text {vap }}{ }^{\circ} \text { for } 1 \mathrm{~mole}=30.5 \mathrm{~kJ} / \mathrm{mol} \\
& \Delta \mathrm{H}_{\text {vap }}{ }^{\circ} \text { for } 1.844 \mathrm{~mol}=30.5 \times 1.844 \\
& \quad=56.242 \mathrm{~kJ}
\end{aligned}$$
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