Some Basic Concepts of Chemistry 1 Question 19
40. The mole fraction of a solute in a solution is 0.1 . At $298 \mathrm{~K}$, molarity of this solution is the same as its molality. Density of this solution at $298 \mathrm{~K}$ is $2.0 \mathrm{~g} \mathrm{~cm}^{-3}$. The ratio of the molecular weights of the solute and solvent, $\left(\frac{m_{\text {solute }}}{m_{\text {solvent }}}\right)$ is …
(2016 Adv.)
Show Answer
Solution:
- Moles of solute, $n_{1}=\frac{w_{1}}{m_{1}}$; Moles of solvent, $n_{2}=\frac{w_{2}}{m_{2}}$
$$ \chi_{1}(\text { solute })=0.1 \text { and } \chi_{2}(\text { solvent })=0.9 $$
$$ \therefore \quad \frac{\chi_{1}}{\chi_{2}}=\frac{n_{1}}{n_{2}}=\frac{w_{1}}{m_{1}} \cdot \frac{m_{2}}{w_{2}}=\frac{1}{9} $$
Molarity $=\frac{\text { Solute }(\text { moles })}{\text { Volume }(\mathrm{L})}=\frac{w_{1} \times 1000 \times 2}{m_{1}\left(w_{1}+w_{2}\right)}$
Note Volume $=\frac{\text { Total mass of solution }}{\text { Density }}=\left(\frac{w_{1}+w_{2}}{2}\right) \mathrm{mL}$
Molality $=\frac{\text { Solute }(\text { moles })}{\text { Solvent }(\mathrm{kg})}=\frac{w_{1} \times 1000}{m_{1} \times w_{2}}$
Given,
molarity $=$ molality
hence,
$$ \frac{2000 w_{1}}{m_{1}\left(w_{1}+w_{2}\right)}=\frac{1000 w_{1}}{m_{1} w_{2}} $$
$\therefore \quad \frac{w_{2}}{w_{1}+w_{2}}=\frac{1}{2} \Rightarrow w_{1}=w_{2}=1$
$\therefore \quad \frac{w_{1} m_{2}}{m_{1} w_{2}}=\frac{1}{9} \Rightarrow \frac{m_{1}(\text { solute })}{m_{2} \text { (solvent) }}=9$
