1.

Calculate the ebullioscopic constant for water. The heat of vaporization is 40.685 kJ mol^(-1)

Answer»

`0.512 K KG mol^(-1)`
`1.86 K kg mol^(-1)`
`5.12 K kg mol^(-1)`
`3.56 Kkg mol^(-1)`

SOLUTION :`K_(b)=(RT_(0)^(2)M)/(1000DeltaH_("vapour"))`
`=(8.314xx(373.15)^(2)xx18)/(1000xx40.685)xx10^(-3)=0.512 K kg mol^(-1)`


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