An ideal solution was prepared by dissolving some amount of cane sugar (non-volatile) in `0.9` moles of water. The solution was then cooled just below its freezing temperature `(271K)`, where some ice get separated out. The remaining aqueous solution registered a vapour pressure of 700 torr at `373 K`. Calculate the mass of ice separated out, if the molar heat of fusion of water is `96 kJ`.

Answer

`12.54g`

Explanation

Initial mole of `H_(2)O = 0.9`
`DeltaT_(f) = 6 kJ`
`k_(f) = (RT_(f)^(2)M)/(1000 DeltaH_(f)) = (8.314 xx (273)^(2)xx18)/(1000 xx 6000) = 1.86`
`DeltaT_(f) = k_(f) xx m`
`m = (2)/(1.86) = 1.075`
so in `1000g H_(2)O rarr 1.075` mole solute
in `1g H_(2)O rarr (1.075)/(1000)` mole solute
in `0.9 xx 18 g H_(2)O rarr (1.075)/(1000) xx 0.9 xx 18` mole solute
mole of solute `(n) = 0.0174.15`
`(P^(@)-P_(s))/(P_(s)) = (n)/(N) = (760-700)/(0.0851) = 0.0857`
mole of `H_(2)O(N) = (0.017415)/(0.0857) = 0.2032`
mole of Ice separate out `= 0.9 - 0.2032 = 0.6963`
mass of Ice separate out `= 0.6968 xx 18 = 12.54g`