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

12.54

Explanation

`Delta"T"_("f")=("RT"_("f")^(2)"m"_("solvent"))/(Delta_("f")"H"xx1000)xx("n"_("solute")xx1000)/("W"_("solvent"))`
`2=(8.314xx(273)^(2)xx18)/(60000xx1000)xx("n"_("solute")xx1000)/("W"_("solvent"))`
`"n"_("solute")=0.0174`
Let 'W' gm of ice seperated out
`("P"_("solvent")^(0)-"P"_("solution"))/("P"_("solution"))=("n"_("solute"))/("n"_("solvent"))`
`(760-700)/(700)=(0.0174)/(((0.9xx18-"w"))/(18))`
`"w"=12.54"gm"`