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Choose the Best Answer · Q23

Q.Molar heat of vapourisation of a liquid is 4.8 kJ mol−1^{-1}. If the entropy change is 16 J mol−1^{-1} K−1^{-1}, the boiling point of the liquid is

(a) 323 K
(b) 27∘27^\circC
(c) 164 K
(d) 0.3 K
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Step 1. At the boiling point, ΔSv=ΔHv/Tb\Delta S_v=\Delta H_v/T_b (Section ~7.10.4, eq. 7.33), so Tb=ΔHv/ΔSvT_b=\Delta H_v/\Delta S_v.

Step 2. ΔHv=4.8kJ mol−1=4800J mol−1\Delta H_v=4.8\\ \text{kJ mol}^{-1}=4800\\ \text{J mol}^{-1}; ΔSv=16J mol−1K−1\Delta S_v=16\\ \text{J mol}^{-1}\text{K}^{-1}. …

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