If the freezing point of a 5% w/w aqueous so | vedclass

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(A) The depression in freezing point is given by $\Delta T_f = T_f^\circ - T_f$. For sucrose ($C_{12}H_{22}O_{11}$,molar mass $M_1 = 342 \ g/mol$): $\

Vedclass · Accepted Answer

$269.07$

Vedclass · Answer

(A) The depression in freezing point is given by $\Delta T_f = T_f^\circ - T_f$. For sucrose ($C_{12}H_{22}O_{11}$,molar mass $M_1 = 342 \ g/mol$): $\Delta T_f = 273.15 \ K - 271 \ K = 2.15 \ K$. Since $\Delta T_f = K_f 	imes m$,where $m = \frac{w_2 	imes 1000}{M_2 	imes w_1}$,for a $5\% \ w/w$ solution,$m = \frac{5 	imes 1000}{M_2 	imes 95}$. Thus,$\Delta T_f \propto \frac{1}{M_2}$. For glucose ($C_6H_{12}O_6$,molar mass $M_2 = 180 \ g/mol$): $\frac{\Delta T_{f, 	ext{glucose}}}{\Delta T_{f, 	ext{sucrose}}} = \frac{M_{1}}{M_{2}} = \frac{342}{180} = 1.9$. $\Delta T_{f, 	ext{glucose}} = 1.9 	imes 2.15 \ K = 4.085 \ K$. Therefore,$T_{f, 	ext{glucose}} = 273.15 \ K - 4.085 \ K = 269.065 \ K \approx 269.07 \ K$.

If the freezing point of a $5\% \ w/w$ aqueous solution of sucrose is $271 \ K$ and pure water has a freezing point of $273.15 \ K$,then calculate the freezing point of a $5\% \ w/w$ aqueous solution of glucose. (in $K$)

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