The mole fraction of a solute in a solution is $0.1$ . At $298 K$ , molarity of this solution is the same as its molality. Density of this solution at $298 K$ is $2.0 {g cm}^{- 3}$ . The ratio of the molecular weights of the solute and solvent, $(\frac{m_{solute}}{m_{solvent}})$ is

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Solution

Moles of solute, $n_{1} = \frac{w_{1}}{m_{1}}$
where, $w_{1} = mass of solute in gram$
$m_{1} = molar mass of solute$
Moles of solvent, $n_{2} = \frac{w_{2}}{m_{2}}$
where, $w_{2} = mass of solvent in gram$
$m_{1} = molar mass of solvent$
$χ_{1} (solute) = 0.1 χ_{2} (solvent) = 0.9$
$∴ \frac{χ_{1}}{χ_{2}} = \frac{n_{1}}{n_{2}} = \frac{w_{1}}{m_{1}} \times \frac{m_{2}}{w_{2}} = \frac{1}{9}$
Volume of the solution $= \frac{Mass of the solution}{D e n s i t y} = \frac{w_{1} + w_{2}}{2} mL$
$Molarity = \frac{Number of moles of solute}{Volume (L)} = \frac{w_{1} \times 1000 \times 2}{m_{1} (w_{1} + w_{2})}$
$Molality = \frac{Number of moles of solute}{Mass of solvent (kg)} = \frac{w_{1} \times 1000}{m_{1} \times w_{2}}$
Given that $Molarity = molality \Rightarrow \frac{w_{1} \times 1000 \times 2}{m_{1} (w_{1} + w_{2})} = \frac{w_{1} \times 1000}{m_{1} \times w_{2}} \Rightarrow \frac{w_{2}}{w_{1} + w_{2}} = \frac{1}{2} \Rightarrow w_{1} = w_{2}$
$∴ \frac{w_{1}}{m_{1}} \times \frac{m_{2}}{w_{2}} = \frac{1}{9} \Rightarrow \frac{m_{1}}{m_{2}} = 9$

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