The molal depression constant for water is 1. | vedclass

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Question

(A) The molar mass of sugar $(C_{12}H_{22}O_{11})$ is $12 	imes 12 + 22 	imes 1 + 11 	imes 16 = 144 + 22 + 176 = 342\,g/mol$.Given mass of solute $

Vedclass · Accepted Answer

$-1.86$

Vedclass · Answer

(A) The molar mass of sugar $(C_{12}H_{22}O_{11})$ is $12 	imes 12 + 22 	imes 1 + 11 	imes 16 = 144 + 22 + 176 = 342\,g/mol$.Given mass of solute $(w_2) = 342\,g$.Given mass of solvent $(w_1) = 1000\,g = 1\,kg$.Molality $(m) = \frac{	ext{moles of solute}}{	ext{mass of solvent in kg}} = \frac{342/342}{1} = 1\,m$.The depression in freezing point $(\Delta T_f)$ is given by $\Delta T_f = K_f 	imes m = 1.86 	imes 1 = 1.86\,^{\circ}C$.Freezing point of solution $(T_f) = T_f^{\circ} - \Delta T_f = 0 - 1.86 = -1.86\,^{\circ}C$.

The molal depression constant for water is $1.86\,^{\circ}C/m$. If $342\,g$ of sugar $(C_{12}H_{22}O_{11})$ is dissolved in $1000\,g$ of water,the freezing point of the solution will be ............. $^{\circ}C$.

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