1.

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


Discussion

No Comment Found

Related InterviewSolutions