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(C) The formula for depression in freezing point is $\Delta T_f = K_f 	imes m$,where $m$ is the molality.Molality $m = \frac{w_2 	imes 1000}{M_2 	i

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$3$

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(C) The formula for depression in freezing point is $\Delta T_f = K_f 	imes m$,where $m$ is the molality.Molality $m = \frac{w_2 	imes 1000}{M_2 	imes w_1}$,where $w_2$ is the mass of solute (urea),$M_2$ is the molar mass of urea $(60 \ g/mol)$,and $w_1$ is the mass of solvent (water).Given: $\Delta T_f = 0.186 \ ^oC$,$K_f = 1.86 \ ^oC/m$,$w_1 = 500 \ g$ (since density of water is $1 \ g/mL$,$500 \ mL = 500 \ g$).Substituting the values: $0.186 = 1.86 	imes \frac{w_2 	imes 1000}{60 	imes 500}$.$0.186 = 1.86 	imes \frac{w_2 	imes 2}{60}$.$0.186 = 1.86 	imes \frac{w_2}{30}$.$w_2 = \frac{0.186 	imes 30}{1.86} = 0.1 	imes 30 = 3 \ g$.

The mass of urea required to be dissolved in $500 \ mL$ of water to produce a depression in freezing point of $0.186 \ ^oC$ is ........ $g$. $(K_f = 1.86 \ ^oC/m)$

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