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(C) Given: Mass of solute $(W_2)$ = $15 \ g$,Volume of water = $200 \ mL$,so Mass of solvent $(W_1)$ = $200 \ g$ (assuming density = $1 \ g/mL$).Depre

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$186 \ g \ mol^{-1}$

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(C) Given: Mass of solute $(W_2)$ = $15 \ g$,Volume of water = $200 \ mL$,so Mass of solvent $(W_1)$ = $200 \ g$ (assuming density = $1 \ g/mL$).Depression in freezing point $(\Delta T_f)$ = $0.75 \ K$.Cryoscopic constant $(K_f)$ = $1.86 \ K \ kg \ mol^{-1}$.The formula for molar mass $(M_2)$ is:$M_2 = \frac{1000 	imes K_f 	imes W_2}{\Delta T_f 	imes W_1}$Substituting the values:$M_2 = \frac{1000 	imes 1.86 	imes 15}{0.75 	imes 200}$$M_2 = \frac{27900}{150} = 186 \ g \ mol^{-1}$.

$A$ solution of a nonvolatile solute is obtained by dissolving $15 \ g$ in $200 \ mL$ of water,which has a depression in freezing point of $0.75 \ K$. Calculate the molar mass of the solute if the cryoscopic constant of water is $1.86 \ K \ kg \ mol^{-1}$.

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