5 g sucrose (molar mass = 342 g mol^{-1}) | vedclass

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Question

(A) The formula for depression in freezing point is $\Delta T_{f} = K_{f} \cdot m$. First,calculate the molality $(m)$ of the solution: $m = \frac{	e

Vedclass · Accepted Answer

$14.7 \ K \ kg \ mol^{-1}$

Vedclass · Answer

(A) The formula for depression in freezing point is $\Delta T_{f} = K_{f} \cdot m$. First,calculate the molality $(m)$ of the solution: $m = \frac{	ext{moles of solute}}{	ext{mass of solvent in kg}} = \frac{5 / 342}{0.1 \ kg} = \frac{50}{342} \ mol \ kg^{-1}$. Now,substitute the values into the depression in freezing point equation: $2.15 = K_{f} \cdot \frac{50}{342}$. Solving for $K_{f}$: $K_{f} = \frac{2.15 	imes 342}{50} = 14.7 \ K \ kg \ mol^{-1}$.

$5 \ g$ sucrose (molar mass $= 342 \ g \ mol^{-1}$) is dissolved in $100 \ g$ of solvent,decreasing the freezing point by $2.15 \ K$. What is the cryoscopic constant $(K_{f})$ of the solvent?

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