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(NONE) The depression in freezing point is given by the formula: $\Delta T_{f} = K_{f} 	imes m$,where $m$ is the molality of the solution.Given: $\De

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$15$

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(NONE) The depression in freezing point is given by the formula: $\Delta T_{f} = K_{f} 	imes m$,where $m$ is the molality of the solution.Given: $\Delta T_{f} = 0 - (-24) = 24 \ K$,$K_{f} = 1.86 \ K \ kg \ mol^{-1}$,Mass of solvent $(W_{solvent})$ $= 18.6 \ kg$,Molar mass of solute $(M_{solute})$ $= 62 \ g \ mol^{-1}$.Molality $(m)$ $= \frac{W_{solute} \ (g)}{M_{solute} 	imes W_{solvent} \ (kg)} = \frac{W_{solute}}{62 	imes 18.6}$.Substituting the values: $24 = 1.86 	imes \frac{W_{solute}}{62 	imes 18.6}$.$24 = \frac{1.86 	imes W_{solute}}{1153.2}$.$W_{solute} = \frac{24 	imes 1153.2}{1.86} = 14880 \ g$.Converting to $kg$: $W_{solute} = 14.88 \ kg$.

Mass of ethylene glycol (antifreeze) to be added to $18.6 \ kg$ of water to protect the freezing point at $-24^{\circ} C$ is . . . . . . $kg$ (Molar mass in $g \ mol^{-1}$ for ethylene glycol $= 62$,$K_{f}$ of water $= 1.86 \ K \ kg \ mol^{-1}$)

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