The latent heat of fusion of ice is 80 calories per gram at `0^(@)C`. What is the freezing point of a solution of `KCl` in water containing `7.45` grams of solute 500 grams of water, assuming that the salt is dissociated to the extent of `95%`?

Answer

`T_(f) =- 0.73^(@)C`

Explanation

`(K_(f))H_(2)O =(RT_(f)^(2)M_("solvent"))/(DeltaH_("fusion",mxx1000)`
`=(2xx273.15^(2)xx18)/(80 xx 18 xx 1000) = (2xx74610.9)/(80xx1000)`
`= 1.865 K-kg//mol`
`T'_(f) = 0^(@)C -ik_(f).m`
`= 0^(@)C -(1.95)(1.865)(7.45//74.5)/(0.5) =- 0.73^(@)C`