Ethylene glycol is used as an antifreeze in c | vedclass

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(C) 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.Here,$\Delta T_f =

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$800.00$

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(C) 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.Here,$\Delta T_f = 0 - (-6) = 6 \ K$.$K_f = 1.86 \ K \ kg \ mol^{-1}$.Mass of solvent $(W_A)$ $= 4 \ kg$.Molar mass of solute $(M_B)$ $= 62 \ g \ mol^{-1}$.Molality $(m)$ $= \frac{W_B}{M_B 	imes W_A (	ext{in } kg)}$.Substituting the values: $6 = 1.86 	imes \frac{W_B}{62 	imes 4}$.$W_B = \frac{6 	imes 62 	imes 4}{1.86}$.$W_B = \frac{1488}{1.86} = 800 \ g$.

Ethylene glycol is used as an antifreeze in cold climates. To prevent $4 \ kg$ of water from freezing at $-6^{\circ}C$,how many grams of ethylene glycol must be added? (Given for water: $K_f = 1.86 \ K \ kg \ mol^{-1}$,Molar mass of ethylene glycol $= 62 \ g \ mol^{-1}$)

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