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

Let m. be the molality of the solution after the ice separates out at `-3.534^(@)C`
Now we have,
`DeltaT_(f)=K_(f)*m.`
`:.m.=(DeltaT_(f))/(K_(f))=(3.534)/(1.86)=1.9`
Let us now calculate the amount of ice separated.
Initially the molality is 1m and wt. of solution is 1000g
1 mole of sucrose is dissolved in 1000g of `H_(2)O`
or 342g of sucrose is dissolved in 1000g of `H_(2)O`
`:.1342g` of solution contained `342g` of sucrose
`:.1000g` of solution contained `(342)/(1342)xx1000=254.84g`
Amount of `H_(2)O=1000-254.84=745.16g`
Now, when ice separates out, the molality is 1.9 and the weight of sucrose remains the same as before
`:.(1.9xx342)`g of sucrose is present in 1000g of `H_(2)O`.
`:.254.84g` of sucrose should be in `(1000xx254.84)/(1.9xx342)`
=392.18g of `H_(2)O`
Thus amount of ice separated `=745.16-392.18`
`=352.98g`