    Question

# 1000 g of 1 m aqueous solution of sucrose is cooled and maintained at −3.534∘C. What amount of ice will separate out at this temperature? Kf(H2O)=1.86 K⋅kg⋅mol−1.

A
​​​​​​451 g
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B
​​​​​​301 g
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C
​​​​​​400 g
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D
​​​​​​353 g
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Solution

## The correct option is D ​​​​​​353 gFreezing point depression of a solution is given by, △Tf=Kf×m where, △Tf is the depression in freezing point. Kf is molal depression constant. m is molality of solution. Let m′ be the molality of the solution after the ice separates out at −3.534oC. Thus, we have △Tf=Kf⋅m′ ∴ m′=△TfKf=3.5341.86=1.9 Let us now calculate the amount of ice separated. Initially the molality of solution is 1 m and weight of solution is 1000 g 1 mol of sucrose is dissolved in 1000 g of H2O or 342 g of sucrose is dissolved in 1000 g of H2O. ∴ 1342 g of solution contained 342 g of sucrose. ∴1000 g of solution contained 3421342×1000=254.84 g. Amount of H2O=1000−254.84=745.16 g Now, when ice separates out, the molality is 1.9 m and the weight of sucrose remains the same as before. ∵(1.9×342) g of sucrose is present in 1000 g of H2O ∴254.84 g of sucrose should be in 1000×254.841.9×342 =392.18 g of H2O Thus, the amount of ice separated =745.16−392.18 =352.98=353 g  Suggest Corrections  1      Similar questions  Explore more