An ideal solution was prepared by dissolving some amount of can sugar (non-volatile) in 0.9 moles of water.The solution was then cooled just below its freezing temperature (271 K) where some ice get separated out.The remaining aqueous solution registered a vapour pressure of 700 torr at 373K.Calculate the mass of ice separated out, if the molar heat of fusion of water is 6 kJ.

An ideal solution was prepared by dissolving some amount of can sugar

1.	An ideal solution was prepared by dissolving some amount of can sugar (non-volatile) in 0.9 moles of water.The solution was then cooled just below its freezing temperature (271 K) where some ice get separated out.The remaining aqueous solution registered a vapour pressure of 700 torr at 373K.Calculate the mass of ice separated out, if the molar heat of fusion of water is 6 kJ.
Answer» Solution :`K_f=(RT_f^2)/(DeltaH_("fusion"))M`(M=mol. WT. `T_f` =normalfreezing pt) `K_f=(8.314xx(273)^2xx18)/(1000xx6xx10^3)` `:. K_f ~~1.86K.kg mol^(-1)` Now `DeltaT_f=K_fxxm`(m=molality) `:.m=(DeltaT_f)/K_f=((273-271))/1.86=1.07` MOLES/kg But `m=("moles of solutes")/("weight of solvent (in Kg)")` n=moles of solute =`1.075xx((0.9xx18)/1000)=1.74xx10^(-2)` ALSO `X_("solute")=(760-700)/760=0.079` (where X=mole fraction) `:. "Total moles"=(1.74xx10^(-2))/0.079=0.22` Moles of solvent `(H_2O)=0.2026` `:.` Mass of ice SEPARATED out =(0.9-0.2026)x18=12 gm