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Q.
A solution of x moles of sucrose in 100 grams of water freezes at -0.2∘C As ice separates the freezing point goes down to 0.25∘C How many grams of ice would have
separated?
Solutions
Solution:
(i)ΔTf=m×Kf 0.2=X×1000100×1.86 X=0.210×1.86
after freezing ΔTf=m×Kf ΔTf=X×1000(100−y)×1.86 ΔTf=0.25
On solving, Amount of ice y = 20 g ice