What should be the freezing point of an aqueo | vedclass

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(A) The molar mass of $C_2H_5OH$ is $(2 	imes 12) + (6 	imes 1) + 16 = 46 \ g \ mol^{-1}$.The molality $(m)$ of the solution is calculated as: $m =

Vedclass · Accepted Answer

$- 0.69$

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(A) The molar mass of $C_2H_5OH$ is $(2 	imes 12) + (6 	imes 1) + 16 = 46 \ g \ mol^{-1}$.The molality $(m)$ of the solution is calculated as: $m = \frac{	ext{mass of solute}}{	ext{molar mass of solute} 	imes 	ext{mass of solvent in kg}} = \frac{17 \ g}{46 \ g \ mol^{-1} 	imes 1 \ kg} = 0.3696 \ mol \ kg^{-1}$.The depression in freezing point is given by: $\Delta T_f = K_f 	imes m = 1.86 \ K \ kg \ mol^{-1} 	imes 0.3696 \ mol \ kg^{-1} \approx 0.6875 \ ^oC \approx 0.69 \ ^oC$.The freezing point of the solution is $T_f = T_f^{\circ} - \Delta T_f = 0 \ ^oC - 0.69 \ ^oC = - 0.69 \ ^oC$.

What should be the freezing point of an aqueous solution containing $17 \ g$ of $C_2H_5OH$ in $1000 \ g$ of water? (Given: $K_f$ of water = $1.86 \ K \ kg \ mol^{-1}$)

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