Standard enthalpy of vapourisation for CCl 4 is 30.5 kJ mol -1 . Heat required for vapourisation of 284 g g of CCl 4 at constant temperature is kJ. (Given molar mass in g mol -1 ; C =12, Cl =35.5 )

Chemistry · Thermodynamics & Thermochemistry

JEE Main 2024 — 29 January, Shift 2 — Question 82

Standard enthalpy of vapourisation for $\mathrm{CCl}_{4}$ is $30.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$ .

Heat required for vapourisation of $284 \mathrm{~g}_{\mathrm{g}}$ of $\mathrm{CCl}_{4}$ at constant temperature is $\qquad$ kJ.

(Given molar mass in $\mathrm{g} \mathrm{mol}^{-1} ; \mathrm{C}=12, \mathrm{Cl}=35.5$ )

Numerical answer — enter this value.

$\Delta \mathrm{H}_{\text {vap }}^{0} \mathrm{CCl}_{4}=30.5 \mathrm{~kJ} / \mathrm{mol}$

Mass of $\mathrm{CCl}_{4}=284 \mathrm{gm}$

Molar mass of $\mathrm{CCl}_{4}=154 \mathrm{~g} / \mathrm{mol}$

Moles of $\mathrm{CCl}_{4}=\frac{284}{154}=1.844 \mathrm{~mol}$

$\Delta \mathrm{H}_{\text {vap }}{ }^{\circ}$ for $1 \mathrm{~mole}=30.5 \mathrm{~kJ} / \mathrm{mol}$

$\Delta \mathrm{H}_{\text {vap }}{ }^{\mathrm{o}}$ for $1.844 \mathrm{~mol}=30.5 \times 1.844$ $=56.242 \mathrm{~kJ}$

Answer key and solution verified before publishing.

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