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Question

The freezing point of aqueous solution that contains 3% urea, 7.45% KCl and 9% of glucose is: (given K1 of water =1.86 and assume molarity=molality).

A
290 K
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B
285.5 K
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C
267.42 K
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D
250 K
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Solution

The correct option is B 267.42 K
urea = 3%
KCl = 7.45%
Glucose = 9%
Let Amount of water be 100gmNH2CONH2 urea = 3gm molar mass = 60
KCl=7.45g molar mass = 74.5 C6H12O6 Glucose = 9g Molar mass = 180
total no of mole = 360+7.4574.5+9180
120+110+120
=15=0.2
molality = 0.2100=2m
ΔTf=kfm
=1.86×2=3.72
Freezing point = 2733.72
=269.28k

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