1000 g of 1 molal aqueous solution of sucrose is cooled and maintained at `-3.534^(@)C`. Find out how much ice will separate out at this temperature. `(K_(f)` for water `=1.86 k m^(-1)`)

Explanation

`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`