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(A) The formula for freezing point depression is $\Delta T_f = K_f 	imes m$,where $m$ is the molality of the solution. The mass of the solvent $(W_A)

Vedclass · Accepted Answer

$125 \ g \ mol^{-1}$

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(A) The formula for freezing point depression is $\Delta T_f = K_f 	imes m$,where $m$ is the molality of the solution. The mass of the solvent $(W_A)$ is calculated as: $W_A = 	ext{density} 	imes 	ext{volume} = 0.8 \ g \ mL^{-1} 	imes 100 \ mL = 80 \ g$. Substituting the given values into the formula: $\Delta T_f = K_f 	imes \frac{W_B}{M_B} 	imes \frac{1000}{W_A(g)}$. $0.75 = 5 	imes \frac{1.5}{M_B} 	imes \frac{1000}{80}$. $0.75 = 5 	imes \frac{1.5}{M_B} 	imes 12.5$. $M_B = \frac{5 	imes 1.5 	imes 12.5}{0.75} = \frac{93.75}{0.75} = 125 \ g \ mol^{-1}$.

Calculate the molar mass of the solute when $1.5 \ g$ of a non-volatile solute is dissolved in $100 \ mL$ of a solvent having a density of $0.8 \ g \ mL^{-1}$,which lowers its freezing point by $0.75 \ K$. (Freezing point depression constant for the solvent is $5 \ K \ kg \ mol^{-1}$).

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