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Q.

1 g of a non-volatile, non-electrolyte solute of molar mass 250 g mol- was dissolved in 51.2 g of benzene. If the freezing point depression constant Kf of benzene is 5.12 kg K mol, the freezing point of benzene is lowered by :

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a

0.3 K

b

0.5 K

c

0.2 K

d

0.4 K

answer is D.

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Detailed Solution

Tf=Kf×W1×1000W2×M1 W1=weight of the solute W2=weight of the solvent M1=molar mass of the solute Kf=freezing point depression constant Now, Tf=5.12×1×100051.2×250 Tf=0.4 K

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