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QJEEM24Chff6f: JEE Main 2024 Chemistry - Class 11 JEEM24Chff6f Solution: Standard enthalpy of vapourisation for is . Heat required for…

Standard enthalpy of vapourisation for CCl4\mathrm{CCl}_4 is 30.5 kJ mol−130.5 \mathrm{~kJ} \mathrm{~mol}^{-1} . Heat required for vapourisation of 284 g284 \mathrm{~g} of CCl4\mathrm{CCl}_4 at constant temperature is ________ kJ\mathrm{kJ} .

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

Step-by-Step Verified Solution

amp;ΔHvap 0CCl4=30.5 kJ/molamp; Mass of CCl4=284 gmamp; Molar mass of CCl4=154 g/molamp; Moles of CCl4=284154=1.844 molamp;ΔHvap ∘ for 1 mole=30.5 kJ/molamp;ΔHvap ∘ for 1.844 mol=30.5×1.844amp;=56.242 kJ\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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