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(D) The depression in freezing point is given by the formula: $\Delta T_f = K_f 	imes m$ Here,$m$ is the molality of the solution. The molar mass of

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(D) The depression in freezing point is given by the formula: $\Delta T_f = K_f 	imes m$ Here,$m$ is the molality of the solution. The molar mass of ethanol $(C_2H_5OH)$ is $M_B = (2 	imes 12) + (6 	imes 1) + 16 = 46 \ g \ mol^{-1}$. The mass of solute $(w_B)$ is $25 \ g$ and the mass of solvent $(w_A)$ is $1000 \ g$. $\Delta T_f = K_f 	imes \frac{w_B 	imes 1000}{M_B 	imes w_A} = 1.86 	imes \frac{25 	imes 1000}{46 	imes 1000} = 1.86 	imes 0.543 \approx 1.01^{\circ} C$. The freezing point of the solution is $T_f = T_f^{\circ} - \Delta T_f = 0^{\circ} C - 1.01^{\circ} C = -1.01^{\circ} C$. Rounding to the nearest integer,the freezing point is $-1^{\circ} C$.

The freezing point of an aqueous solution containing $25 \ g$ of ethanol $(C_2H_5OH)$ in $1000 \ g$ of $H_2O$ is $(K_f = 1.86 \ K \ kg \ mol^{-1})$ (in $^{\circ} C$)

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