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Q.

1.00 g of a non-electrolyte solute is dissolved in 50 g of benzene which lowers the freezing point of benzene by 0.40 K. The freezing point of depression constant of benzene is 5.12 K kg mol-1 . Find the molar mass of the solute.

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a

206 g mol-1

b

226 g mol-1

c

246 g mol-1

d

253 g mol-1

answer is D.

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Detailed Solution

M2=Kf×w2×1000ΔTf×w1 where, w1→mass of solvent, w2 →mass of solute and M2→molar mass of solute.Substituting the values of various terms involved in equationM2=5.12Kkgmol−1×1.00g×1000gkg−10.40K×50g=256gmol−1Thus, molar mass of the solute = 256 g mol-1
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1.00 g of a non-electrolyte solute is dissolved in 50 g of benzene which lowers the freezing point of benzene by 0.40 K. The freezing point of depression constant of benzene is 5.12 K kg mol-1 . Find the molar mass of the solute.