1000 g of 1 m sucrose solution in water is cooled to -3.354^(@)C. What mass of ice would separated out at this temperature ? K_(f) for H_(2)O= 1.86" K mol"^(-1)kg.

1000 g of 1 m sucrose solution in water is cooled to -3.354^(@)C. What

1.	1000 g of 1 m sucrose solution in water is cooled to -3.354^(@)C. What mass of ice would separated out at this temperature ? K_(f) for H_(2)O= 1.86" K mol"^(-1)kg.
Answer» Solution :`DeltaT_(f)=K_(f)xxm = 1.86 xx1=1.86^(@)` `therefore ` Freezing point of the solution `=-1.86^(@)C` As TOTAL mass of solution = 1000 G `therefore""w_(1)+w_(2)=1000"...(i)"` Further, `""DeltaT_(f)=(1000K_(f)w_(2))/(w_(1)xxM_(2))` `1.86xx(1000xx1.86xxw_(2))/(w_(1)xx342)"or"(w_(2))/(w_(1))=(1.86xx342)/(1000xx1.86)=0.342"...(ii)"` From eqns. (i)and (ii), on solving, we get`""w_(2)=254.84g` `""w_(1)=745.16g` These were the amounts present in the original solution UPTO -`1.86^(@)C`. Now on further cooling to `-3.534^(@)C`, mass of sucrose will remain the same but some water will freeze out as ICE. The mass of water left can be CALCULATED as follows : `DeltaT_(f)=(1000K_(f)w_(2))/(w_(1)xxM_(2))` `3.534=(1000xx1.86xx254.84)/(w_(1)xx342) or w_(1)=392.18 g` `therefore ` Amount of water separated as ice `=745.16-392.18 g = 352.98 g`