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Calculatte the molal elevation constant of water if molar enthalpy of vapourisation of water at 373K is 40.585 kJ `mol^(-1)`. |
Answer» `K_(b)=(MR(T_(b)^(@))^(2))/((DeltaH_(VAP))),DeltaH_((VAP))=40.585 kJ mol^(-1)=40585 J mol^(-1)` `R=8.314 jK^(-1)mol^(-1),T_(b)^(@)=373K,18 g mol^(-01)=0.018 kg mol^(-1)` `K_(b)=((0.018 kg mol^(-1))xx(8.314JK^(-1)mil^(-1)xx(373K)^(2)))/((40585 J mol^(-1)))` = 0.513 K kg `mol^(-1)`=0.513 K/m |
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