JEE Main 2021 — Solutions Question with Solution
From: JEE Main 2021 (Online) 27th August Morning Shift
Question
1 kg of 0.75 molal aqueous solution of sucrose can be cooled up to 4C before freezing. The amount of ice (in g) that will be separated out is __________. (Nearest integer)
[Given : Kf(H2O) = 1.86 K kg mol1]
[Given : Kf(H2O) = 1.86 K kg mol1]
Enter your answer
Show full solutionCorrect answer: 518.3
Correct answer
518.3
Step-by-step explanation
Let mass of water initially present = x gm
Mass of sucrose = (1000 x) gm
moles of sucrose =
256.5x = 106 1000x
x = 795.86 gm
moles of sucrose = 0.5969
New mass of H2O = a kg
a = 0.2775 kg
ice separated = (795.86 277.5) = 518.3 gm
Mass of sucrose = (1000 x) gm
moles of sucrose =
256.5x = 106 1000x
x = 795.86 gm
moles of sucrose = 0.5969
New mass of H2O = a kg
a = 0.2775 kg
ice separated = (795.86 277.5) = 518.3 gm
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