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1.0 g of a non-electrolyte solute (molar mass 250 g mol^–1) was dissolved in 51.2 g of benzene. If the freezing point depression constant of benzene is 5.12 K kg mol^–1, the lowering in freezing point will be

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0.5 K

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0.2 K

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answer is B.

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Detailed Solution

Molar mass of solute(s) = 250g/mole

$\begin{array}{l} {W_s} = 1g\\ {W_o}(Benzene) = 51.2g\\ {K_f} = 5.12KKgmol{e^{ - 1}} \end{array}$

$\Delta T_f=K_fm$

$\Delta T_f=5.12X\frac{1}{250}X\frac{1000}{51.2}$

$\Delta T_f=0.4K$

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