17.0 g of NH3 completely vapourises at $-$33.42$^\circ$C and 1 bar pressure and the enthalpy change in the process is 23.4 kJ mol$-$1. The enthalpy change for the vapourisation of 85 g of NH3 under the same conditions is _________ kJ.
Answer (integer)
117
Solution
Number of moles of $\mathrm{NH}_{3}=5$
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So, required $\Delta \mathrm{H}=5 \times 23.4$
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$=117 \mathrm{~kJ}$
About this question
Subject: Chemistry · Chapter: Thermodynamics · Topic: Zeroth and First Law
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