Calculate the amount of KCl (MW=74.5 g mol^ | vedclass

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(C) The formula for depression in freezing point is $\Delta T_f = i 	imes K_f 	imes m$,where $i$ is the van't Hoff factor,$K_f$ is the cryoscopic co

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$20.0$

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(C) The formula for depression in freezing point is $\Delta T_f = i 	imes K_f 	imes m$,where $i$ is the van't Hoff factor,$K_f$ is the cryoscopic constant,and $m$ is the molality.For $KCl$,which is a strong electrolyte,$i = 2$ (dissociates into $K^+$ and $Cl^-$).Given: $\Delta T_f = 2 \ K$,$K_f = 1.86 \ K \ kg \ mol^{-1}$,$W_{solvent} = 500 \ g = 0.5 \ kg$,$MW_{KCl} = 74.5 \ g \ mol^{-1}$.Molality $m = \frac{W_{solute} / MW_{solute}}{W_{solvent} (kg)} = \frac{W_{KCl} / 74.5}{0.5}$.Substituting the values: $2 = 2 	imes 1.86 	imes \frac{W_{KCl} / 74.5}{0.5}$.$2 = 3.72 	imes \frac{W_{KCl}}{37.25}$.$W_{KCl} = \frac{2 	imes 37.25}{3.72} \approx 20.0 \ g$.

Calculate the amount of $KCl$ $(MW=74.5 \ g \ mol^{-1})$ required to depress the freezing point of $500 \ g$ of water by $2 \ K$ (in $g$)? $(K_f=1.86 \ K \ kg \ mol^{-1})$

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