How much glucose (molecular weight = 180 g/ | vedclass

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(A) The depression in freezing point is given by $\Delta T_f = K_f 	imes m$,where $m$ is the molality of the solution after ice separates. At $-0.5^{

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$9$

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(A) The depression in freezing point is given by $\Delta T_f = K_f 	imes m$,where $m$ is the molality of the solution after ice separates. At $-0.5^{\circ}C$,$\Delta T_f = 0 - (-0.5) = 0.5 \ K$. The mass of the remaining solvent $(x)$ is $200 \ g - 14 \ g = 186 \ g$. Using the formula $\Delta T_f = K_f 	imes \frac{w 	imes 1000}{M 	imes x}$,where $w$ is the mass of glucose and $M = 180 \ g/mol$: $0.5 = 1.86 	imes \frac{w 	imes 1000}{180 	imes 186}$. $w = \frac{0.5 	imes 180 	imes 186}{1.86 	imes 1000}$. $w = \frac{0.5 	imes 180 	imes 186}{1860} = \frac{0.5 	imes 180}{10} = 0.5 	imes 18 = 9 \ g$.

How much glucose $(molecular \ weight = 180 \ g/mol)$ should be added to $200 \ g \ H_2O$ so that when the solution is cooled to $-0.5^{\circ}C$,$14 \ g$ of ice separates out of the solution: [$K_f = 1.86 \ K \ kg/mol$ and melting point of $H_2O = 0^{\circ}C$] (in $g$)

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