The amount of urea to be dissolved in 500 mL | vedclass

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

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$0.3$

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(D) The formula for depression in freezing point is $\Delta T_f = K_f 	imes m$,where $m$ is the molality.Molality $m = \frac{w 	imes 1000}{M 	imes W}$,where $w$ is the mass of solute (urea,$M = 60 \ g/mol$) and $W$ is the mass of solvent ($500 \ g$ of water).Given $\Delta T_f = 0.186 \ ^oC$,$K_f = 18.6 \ K \ kg \ mol^{-1}$,and $W = 500 \ g$.Substituting the values: $0.186 = 18.6 	imes \frac{w 	imes 1000}{60 	imes 500}$.$0.186 = 18.6 	imes \frac{w}{30}$.$0.186 = 0.62 	imes w$.$w = \frac{0.186}{0.62} = 0.3 \ g$.

The amount of urea to be dissolved in $500 \ mL$ of water $(K_f = 18.6 \ K \ kg \ mol^{-1})$ to produce a depression of $0.186 \ ^oC$ in freezing point is $....... \ g$. (Assume density of water = $1 \ g/mL$)

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