Calculate the cryoscopic constant (K_f) when | vedclass

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(D) The formula for freezing point depression is $\Delta T_f = K_f 	imes m$,where $m$ is the molality of the solution.Molality $m = \frac{	ext{moles

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$1.17 \ K \ kg \ mol^{-1}$

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(D) The formula for freezing point depression is $\Delta T_f = K_f 	imes m$,where $m$ is the molality of the solution.Molality $m = \frac{	ext{moles of solute}}{	ext{mass of solvent in kg}} = \frac{0.8 \ g / 64 \ g \ mol^{-1}}{43 \ g / 1000} = \frac{0.0125 \ mol}{0.043 \ kg} \approx 0.2907 \ mol \ kg^{-1}$.Given $\Delta T_f = 0.34 \ K$,we have $0.34 = K_f 	imes 0.2907$.$K_f = \frac{0.34}{0.2907} \approx 1.169 \ K \ kg \ mol^{-1}$.Rounding to two decimal places,$K_f \approx 1.17 \ K \ kg \ mol^{-1}$.

Calculate the cryoscopic constant $(K_f)$ when $0.8 \ g$ of a non-volatile solute with a molar mass of $64 \ g \ mol^{-1}$ is dissolved in $43 \ g$ of a solvent,lowering the freezing point by $0.34 \ K$.

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