1.

1000g of 1m aqueous solution of sucrose is cooled and maintained at -3.534^(@)C. Find how much ice will separate out. K_(f)(H_(2)O)=1.86K*kg*mol^(-1)

Answer»

Solution :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`


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