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(B) Given: Mass of solute $(W_2) = 4 \,g$, Molar mass of solute $(M_2) = 126 \,g \,mol^{-1}$, Volume of water $= 80 \,mL$. Since the density of water

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$0.74$

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(B) Given: Mass of solute $(W_2) = 4 \,g$, Molar mass of solute $(M_2) = 126 \,g \,mol^{-1}$, Volume of water $= 80 \,mL$. Since the density of water is $1 \,g/mL$, the mass of solvent $(W_1) = 80 \,g$. The formula for depression in freezing point is $\Delta T_f = \frac{1000 	imes K_f 	imes W_2}{M_2 	imes W_1}$. Substituting the values: $\Delta T_f = \frac{1000 	imes 1.86 	imes 4}{126 	imes 80}$. $\Delta T_f = \frac{7440}{10080} \approx 0.738 \,K$, which rounds to $0.74 \,K$.

Calculate the depression in freezing point of a solution when $4 \,g$ of a nonvolatile solute with a molar mass of $126 \,g \,mol^{-1}$ is dissolved in $80 \,mL$ of water. $[$Cryoscopic constant of water $K_f = 1.86 \,K \,kg \,mol^{-1}]$ (in $\,K$)

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