What is the cryoscopic constant of water if 5 | vedclass

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(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{m

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$7.74 \ K \ kg \ mol^{-1}$

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(A) The formula for depression in freezing point is $\Delta T_f = K_f \cdot m$. First,calculate the molality $(m)$ of the solution: $m = \frac{	ext{moles of solute}}{	ext{mass of solvent in } kg} = \frac{5 \ g / 180 \ g \ mol^{-1}}{0.1 \ kg} = \frac{5}{18} \ mol \ kg^{-1} \approx 0.2778 \ mol \ kg^{-1}$. Given $\Delta T_f = 2.15 \ K$. Substituting the values into the formula: $2.15 = K_f \cdot (5/18)$. $K_f = \frac{2.15 	imes 18}{5} = 7.74 \ K \ kg \ mol^{-1}$.

What is the cryoscopic constant of water if $5 \ g$ of glucose in $100 \ g$ of water causes a depression in freezing point of $2.15 \ K$? (Molar mass of glucose $= 180 \ g \ mol^{-1}$)

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