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Depression in Freezing Point

1.00 g of a n...

Question

$1.00 g$ of a non-electrolyte solute (molar mass $250 g m o l^{- 1}$ ) was dissolved in $51.2 g$ of benzene. If the freezing point depression constant, $K_{f}$ of benzene is $5.12 K k g m o l^{- 1}$ , the freezing point of benzene will be lowered by:

0.4 K

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0.3 K

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0.5 K

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0.2 K

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Solution

The correct option is C $0.4 K$
Molality m is the ratio of the number of moles of solute to the mass of solvent in kg
$m = \frac{1.00 \times 1000}{250 \times 51.2} = 0.078125 m$
The freezing point of benzene will be lowered by $Δ T_{f} = K_{f} m = 5.12 \times 0.078125 = 0.4 K$

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