If the freezing point of an aqueous urea solu | vedclass

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Question

(A) The depression in freezing point is given by $\Delta T_f = T_f^{\circ} - T_f = 273 \ K - 271.14 \ K = 1.86 \ K$. Using the formula $\Delta T_f = K

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$\frac{1}{56.5}$

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(A) The depression in freezing point is given by $\Delta T_f = T_f^{\circ} - T_f = 273 \ K - 271.14 \ K = 1.86 \ K$. Using the formula $\Delta T_f = K_f 	imes m$,where $m$ is the molality: $1.86 = 1.86 	imes m \implies m = 1 \ mol/kg$. This means $1 \ mol$ of urea is dissolved in $1000 \ g$ of water. Moles of water = $\frac{1000 \ g}{18 \ g/mol} \approx 55.55 \ mol$. Mole fraction of urea $(X_{	ext{urea}})$ = $\frac{n_{	ext{urea}}}{n_{	ext{urea}} + n_{	ext{water}}} = \frac{1}{1 + 55.55} = \frac{1}{56.55} \approx \frac{1}{56.5}$.

If the freezing point of an aqueous urea solution is $271.14 \ K$ at $1 \ \text{atm}$ pressure (given $K_f$ of water = $1.86 \ K \ kg/mol$),then what is the mole fraction of urea in this solution? (Freezing point of pure water is $273 \ K$)

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