Molal depression constant of water is $1.86 K K g m o l^{- 1}$ . $0.02 m o l$ of urea dissolved in $100 g$ of water will freeze at :

{0.186}^{o} C

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- {0.372}^{o} C

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{1.86}^{o} C

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{3.72}^{o} C

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Solution

The correct option is B $- {0.372}^{o} C$
Depression in freezing point is given by

$Δ T_{f} = i \times K_{f} \times m$

$Δ T_{f} = T_{f}^{o} - T_{f}$

$i =$ Vant hoff factor $= 1$ (for non electrolyte)

$K_{f} = 1.86 K / m$

$m =$ Molality
$Δ T_{f} = i \times K_{f} \times \frac{moles of solute}{weight of solvent in kg}$

Moles of solution (urea) $= 0.02$
Molar mass of solute (urea) $= 60 g / m o l$
Weight of solvent (water) $= 100 g = 0.1 k g$

$(T_{f}^{0} - T_{f}) = 1 \times 1.86 \times \frac{0.02}{0.1 k g}$

$(0 - T_{f})^{o} C = 0.372$

$T_{f} = - {0.372}^{o} C$

$∴$ Option B is correct.

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