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Q.

The freezing point of aqueous solution that contains 5% by mass urea, 1.0% by mass KCl and 10% by mass of glucose is

Kf(H2O=1.86k molality-1)

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a

250k

b

290.2k

c

269.93k

d

285.5k

answer is C.

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Detailed Solution

Tfo-Tf=Kf×m

n1=560  n2=174.5, n3=10180

For KCl i = 2 then n2=2×174.5

n=560+274.5+10180

Tfo-Tf=1.86×0.164×1000300

273-Tf=1.0168

∴Tf=271.98

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