The solution containing 6 g urea (molar mass | vedclass

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(A) Since both solutions freeze at the same temperature,their freezing point depression $(\Delta T_f)$ is equal. For dilute aqueous solutions,$\Delta

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$90$

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(A) Since both solutions freeze at the same temperature,their freezing point depression $(\Delta T_f)$ is equal. For dilute aqueous solutions,$\Delta T_f = K_f 	imes m$,where $m$ is the molality. Since $K_f$ is the same for both,$m_1 = m_2$. Given $1 \ dm^3$ of water $\approx 1 \ kg$ of water. Molality $m = \frac{	ext{moles of solute}}{	ext{mass of solvent in kg}}$. For urea: $m_1 = \frac{6 \ g / 60 \ g \cdot mol^{-1}}{1 \ kg} = 0.1 \ mol \cdot kg^{-1}$. For solute $A$: $m_2 = \frac{9 \ g / M_A}{1 \ kg} = \frac{9}{M_A} \ mol \cdot kg^{-1}$. Equating $m_1 = m_2$: $0.1 = \frac{9}{M_A}$. Therefore,$M_A = \frac{9}{0.1} = 90 \ g \cdot mol^{-1}$.

The solution containing $6 \ g$ urea (molar mass $60$) per $dm^3$ of water and another solution containing $9 \ g$ of solute $A$ per $dm^3$ water freeze at the same temperature. What is the molar mass of $A$?

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