Calculate the cryoscopic constant (K_f) of a | vedclass

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(D) The formula for the depression in freezing point is $\Delta T_f = K_f 	imes m$,where $m$ is the molality. Molality $m = \frac{W_2 	imes 1000}{M_

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$4.91 \ K \ kg \ mol^{-1}$

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(D) The formula for the depression in freezing point is $\Delta T_f = K_f 	imes m$,where $m$ is the molality. Molality $m = \frac{W_2 	imes 1000}{M_2 	imes W_1}$,where $W_2$ is the mass of solute,$M_2$ is the molar mass of solute,and $W_1$ is the mass of solvent in grams. Substituting the values: $\Delta T_f = 3 \ K$,$W_2 = 2.5 \ g$,$M_2 = 117 \ g \ mol^{-1}$,$W_1 = 35 \ g$. $K_f = \frac{\Delta T_f 	imes M_2 	imes W_1}{1000 	imes W_2}$ $K_f = \frac{3 	imes 117 	imes 35}{1000 	imes 2.5} \ K \ kg \ mol^{-1}$ $K_f = \frac{12285}{2500} \ K \ kg \ mol^{-1} = 4.91 \ K \ kg \ mol^{-1}$.

Calculate the cryoscopic constant $(K_f)$ of a solvent when $2.5 \ g$ of a solute is dissolved in $35 \ g$ of the solvent,which lowers its freezing point by $3 \ K$. (Molar mass of the solute is $117 \ g \ mol^{-1}$)

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