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Question

By dissolving 13.6 g of a substance in 20 g of water, the freezing point decreased by 3.7∘C. Calculate the molecular mass of the substance. (Molal depression constant for water = 1.863K kg mol−1)

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/molFreezing 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 mol−1w=13.6 g,W=20 g,△Tf=3.7∘CM=1000×1.863×13.620×3.7⇒342.39 Molar mass of the substance is 342.4 g/mol

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