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Molar heat of vaporization of a liquid is `6kJ mol^(-1)`. If the entropy change is `16 J mol^(-1) K^(-1)`, the boiling point of the liquid isA. `375^(@)C`B. 375 KC. 273 KD. `102^(@)C` |
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Answer» Correct Answer - B `DeltaS=16 J mol^(-1) K^(-1)` `T_(b.p.)=(DeltaH_("vapour"))/(DeltaS_("vapour"))=(6xx1000)/(16)=375 K`. |
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