Ethylene glycol is used as an anti-freezing a | vedclass

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(B) 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

Vedclass · Accepted Answer

$804.3$

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(B) 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: $\Delta T_f = 6 \ K$,$K_f = 1.85 \ K \ kg \ mol^{-1}$,$W_A = 4 \ kg$,$M_B = 62 \ g \ mol^{-1}$.Substituting the values in the formula: $\Delta T_f = K_f 	imes \frac{w_B}{M_B 	imes W_A}$$6 = \frac{1.85 	imes w_B}{62 	imes 4}$$w_B = \frac{6 	imes 62 	imes 4}{1.85}$$w_B = \frac{1488}{1.85} \approx 804.32 \ g$Thus,the amount of ethylene glycol required is approximately $804.3 \ g$.

Ethylene glycol is used as an anti-freezing agent. Calculate the amount of ethylene glycol to be added to $4 \ kg$ of water to prevent it from freezing at $-6 \ ^{\circ}C$. ($K_f$ for water $= 1.85 \ K \ kg \ mol^{-1}$) (molar mass of $(CH_2OH)_2 = 62 \ g \ mol^{-1}$) .......... $g$.

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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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