MHT CET · Chemistry · Thermodynamics (C)
1.8 g water is vapourised by supplying 4 kJ heat at \(100^{\circ} \mathrm{C}\). What is the heat of vapourisation of water at same temperature?
- A \(8 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
- B \(40 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
- C \(18 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
- D \(32 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
Answer & Solution
Correct Answer
(B) \(40 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
Step-by-step Solution
Detailed explanation
\(n = \frac{m}{M} = \frac{1.8 \mathrm{~g}}{18 \mathrm{~g/mol}} = 0.1 \mathrm{~mol}\) \(\Delta H_{vap} = \frac{Q}{n} = \frac{4 \mathrm{~kJ}}{0.1 \mathrm{~mol}} = 40 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
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