100 g of sucrose solution in water is colled to -0.5^(@)C. What weight of ice would be separated out at this temperature if solution started to freeze at -0.38^(@)C? K_(f) for H_(2)O="1.86 K kg mol"^(-1).

100 g of sucrose solution in water is colled to -0.5^(@)C. What weight

1.	100 g of sucrose solution in water is colled to -0.5^(@)C. What weight of ice would be separated out at this temperature if solution started to freeze at -0.38^(@)C? K_(f) for H_(2)O="1.86 K kg mol"^(-1).
Answer» Solution :Suppose 100 g of the solution contains `w_(2)` g of the SOLUTE and `w_(1)` g of the solvent. Then `w_(1)+w_(2)=100"…(i)"` `DeltaT_(f)=(1000xxK_(f)xxw_(2))/(w_(1)xxM_(2))` As the solution starts freezing at `-0.38^(@)C`. Hence, when `DeltaT_(f)=0.38^(@)` `0.38=(1000xx1.86xxw_(2))/(w_(1)xx342)"or"(w_(2))/(w_(1))=0.07"...(ii)"` Solving eqns. (i) and (ii), we get `w_(2)=6.6g, w_(1)=93.40 g` Now at `-0.5^(@)C`, some WATER SEPARATES out as ice but solute exists as such, i.e., `w_(2)=6.6g`. Let us caluclate water present `(w_(1))` `0.50=(1000xx1.86xx6.6)/(w_(1)xx342)"or"w_(1)=71.78` `therefore"Weight of iceseparated out "=93.40-71.78=21.62g`