Freezing point of a 4 % aqueous solution of X | vedclass

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(A) The depression in freezing point is given by $\Delta T_f = K_f 	imes m$,where $m$ is the molality of the solution.Since the freezing points are e

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$3$

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(A) The depression in freezing point is given by $\Delta T_f = K_f 	imes m$,where $m$ is the molality of the solution.Since the freezing points are equal,$(\Delta T_f)_X = (\Delta T_f)_Y$,which implies $m_X = m_Y$.For a $4 \%$ aqueous solution of $X$,the mass of $X$ is $4 \ g$ in $96 \ g$ of water. Molality $m_X = \frac{4 	imes 1000}{A 	imes 96}$.For a $12 \%$ aqueous solution of $Y$,the mass of $Y$ is $12 \ g$ in $88 \ g$ of water. Molality $m_Y = \frac{12 	imes 1000}{M_Y 	imes 88}$.Equating the two: $\frac{4}{A 	imes 96} = \frac{12}{M_Y 	imes 88}$.Solving for $M_Y$: $M_Y = \frac{12 	imes 96 	imes A}{4 	imes 88} = \frac{3 	imes 96 	imes A}{88} = \frac{288 	imes A}{88} \approx 3.27 	imes A$.Rounding to the nearest integer,the molecular weight of $Y$ is $3A$.

Freezing point of a $4 \%$ aqueous solution of $X$ is equal to the freezing point of a $12 \%$ aqueous solution of $Y$. If the molecular weight of $X$ is $A$,then the molecular weight of $Y$ is ............. $A$.

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