The molar freezing point constant for water i | vedclass

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Question

(A) The molar mass of cane sugar $(C_{12}H_{22}O_{11})$ is $342\,g\,mol^{-1}$.Given mass of solute $(w_2)$ = $342\,g$,mass of solvent $(w_1)$ = $1000\

Vedclass · Accepted Answer

$ - 1.86$

Vedclass · Answer

(A) The molar mass of cane sugar $(C_{12}H_{22}O_{11})$ is $342\,g\,mol^{-1}$.Given mass of solute $(w_2)$ = $342\,g$,mass of solvent $(w_1)$ = $1000\,g$,and $K_f = 1.86\,^{\circ}C\,kg\,mol^{-1}$.Molality $(m)$ = $\frac{w_2 	imes 1000}{M_2 	imes w_1} = \frac{342 	imes 1000}{342 	imes 1000} = 1\,mol\,kg^{-1}$.Depression in freezing point $(\Delta T_f)$ = $K_f 	imes m = 1.86 	imes 1 = 1.86\,^{\circ}C$.Freezing point of solution $(T_f)$ = $T_f^{\circ} - \Delta T_f = 0\,^{\circ}C - 1.86\,^{\circ}C = - 1.86\,^{\circ}C$.

The molar freezing point constant for water is $1.86\,^{\circ}C\,kg\,mol^{-1}$. If $342\,g$ of cane sugar $(C_{12}H_{22}O_{11})$ are dissolved in $1000\,g$ of water,the solution will freeze at $............\,^{\circ}C$.

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