17.0 g of $\mathrm{NH}_3$ completely vapourises at – $33.42^{\circ} \mathrm{C}$ and 1 bar pressure and the enthalpy change in the process is $23.4 \mathrm{~kJ} \mathrm{~mol}^{-1}$. The enthalpy change for the vapourisation of 85 g of $\mathrm{NH}_3$ under the same conditions is ________ kJ.
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Solution
Given data is for 1 moles and asked for 5 moles so value is 23.4 × 5 = 117 kJ
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