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Question

By dissolving 13.6 g of a substance in 20 g of water, the freezing point decreased by 3.7C. Calculate the molecular mass of the substance. (Molal depression constant for water = 1.863K kg mol1)

A
242.4 g/mol
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B
432.4 g/mol
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C
342.4 g/mol
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D
243.4 g/mol
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Solution

The correct option is C 342.4 g/mol
Freezing 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 the solution.
Molality=TfKf
Molality (m)=wsoluteM×1000Wsolvent

TfKf=wsoluteM×1000Wsolvent
where,
w is weight of solute
W is weight of solvent
M is molar mass of the solute
M=1000Kf×wW×TfGiven, Kf=1.863 K kg mol1w=13.6 g,W=20 g,Tf=3.7CM=1000×1.863×13.620×3.7342.39
Molar mass of the substance is 342.4 g/mol

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