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(B) The depression in freezing point is given by $\Delta T_{f} = K_{f} \cdot m$,where $m$ is the molality of the solution.Given: $\Delta T_{f} = 0 - (

Vedclass · Accepted Answer

$150.5$

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(B) The depression in freezing point is given by $\Delta T_{f} = K_{f} \cdot m$,where $m$ is the molality of the solution.Given: $\Delta T_{f} = 0 - (-0.93) = 0.93\,^{\circ}C$,$K_{f} = 1.86\,^{\circ}C\,kg\,mol^{-1}$,and the solution is $7.0\%$ by mass,meaning $7\,g$ of solute in $93\,g$ of water.Molality $m = \frac{	ext{mass of solute}}{	ext{molar mass of solute} 	imes 	ext{mass of solvent in kg}} = \frac{7}{M 	imes 0.093}$.Substituting the values: $0.93 = 1.86 	imes \frac{7}{M 	imes 0.093}$.$M = \frac{1.86 	imes 7}{0.93 	imes 0.093} = \frac{13.02}{0.08649} \approx 150.5\,g\,mol^{-1}$.

Calculate the molecular weight of a substance whose $7.0\%$ by mass solution in water freezes at $-0.93\,^{\circ}C$. The cryoscopic constant of water is $1.86\,^{\circ}C\,kg\,mol^{-1}$. .......... $g\,mol^{-1}$.

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