Some Basic Concepts of Chemistry - Result Question 102
####40. 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, $\left(\frac{m _{\text {solute }}}{m _{\text {solvent }}}\right)$ is …
(2016 Adv.)
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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 $=\frac{\text { Solute }(\text { moles })}{\text { Volume }(L)}=\frac{w _1 \times 1000 \times 2}{m _1\left(w _1+w _2\right)}$
Note Volume $=\frac{\text { Total mass of solution }}{\text { Density }}=\left(\frac{w _1+w _2}{2}\right) mL$
Molality $=\frac{\text { Solute }(\text { moles })}{\text { Solvent }(kg)}=\frac{w _1 \times 1000}{m _1 \times w _2}$
Given,
molarity $=$ molality
hence,
$$ \frac{2000 w _1}{m _1\left(w _1+w _2\right)}=\frac{1000 w _1}{m _1 w _2} $$
$\therefore \quad \frac{w _2}{w _1+w _2}=\frac{1}{2} \Rightarrow w _1=w _2=1$
$\therefore \quad \frac{w _1 m _2}{m _1 w _2}=\frac{1}{9} \Rightarrow \frac{m _1(\text { solute })}{m _2 \text { (solvent) }}=9$
