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

`Delta"T"_("f")="i"xx"K"_("f")xx"m"`
`[1+("n"-1)alpha]xx("RT"_("f")^(2))/(1000xx"L"_("fus"))xx("W"_("solute")xx1000)/("m"_("solute")xx"W"_("solvent"))`
`[1+(2-1)xx0.95]xx(2xx(273)^(2))/(1000xx80)xx(7.45xx1000)/(74.5xx500)=0.73`
`"T"_("f")=-0.73^(@)"C"`