For water,K_f = 1.86 K kg mol^{-1}. If you | vedclass

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(C) The formula for depression in freezing point is $\Delta T_f = K_f 	imes m$,where $m$ is the molality of the solution.Given: $\Delta T_f = 2.8 \ K

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$93$

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(C) The formula for depression in freezing point is $\Delta T_f = K_f 	imes m$,where $m$ is the molality of the solution.Given: $\Delta T_f = 2.8 \ K$,$K_f = 1.86 \ K \ kg \ mol^{-1}$,mass of solvent $(W_{solvent}) = 1.0 \ kg$,and molar mass of ethylene glycol $(C_2H_6O_2) = 62 \ g \ mol^{-1}$.Substituting the values: $2.8 = 1.86 	imes \frac{W_{solute}}{62 	imes 1.0}$.Solving for $W_{solute}$: $W_{solute} = \frac{2.8 	imes 62}{1.86} \approx 93.33 \ g$.Thus,the required amount is approximately $93 \ g$.

For water,$K_f = 1.86 \ K \ kg \ mol^{-1}$. If your car radiator contains $1.0 \ kg$ of water,how many grams of ethylene glycol $(C_2H_6O_2)$ must be added to lower the freezing point of the solution to $-2.8 \ ^\circ C$?

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