JEE Mains · Chemistry · STD 11 - 5. Thermodynamics and thermochemistry
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 \(\mathrm{g} \mathrm{mol}^{-1} ; \mathrm{C}=12, \mathrm{Cl}=35.5\) )
- A \(78\)
- B \(12\)
- C \(46\)
- D \(56\)
Answer & Solution
Correct Answer
(D) \(56\)
Step-by-step Solution
Detailed explanation
\( \Delta \mathrm{H}_{\text {vap }}^0 \mathrm{CCl}_4=30.5 \mathrm{~kJ} / \mathrm{mol} \) \( \text { Mass of } \mathrm{CCl}_4=284 \mathrm{gm} \) \( \text { Molar mass of } \mathrm{CCl}_4=154 \mathrm{~g} / \mathrm{mol} \)…
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