Some Basic Concepts of Chemistry - Result Question 40
Interger Answer Type Questions
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, $\left(\dfrac{m _{\text {solute }}}{m _{\text {solvent }}}\right)$ is …
(2016 Adv.)
Show Answer
Answer:
Correct Answer: 40. (9)
Solution:
- Moles of solute, $n _1=\dfrac{w _1}{m _1}$; Moles of solvent, $n _2=\dfrac{w _2}{m _2}$
$\chi _1(\text { solute })=0.1 \text { and } \chi _2(\text { solvent })=0.9$
$\therefore \quad \dfrac{\chi _1}{\chi _2}=\dfrac{n _1}{n _2}=\dfrac{w _1}{m _1} \cdot \dfrac{m _2}{w _2}=\dfrac{1}{9}$
Molarity $=\dfrac{\text { Solute }(\text { moles })}{\text { Volume }(L)}=\dfrac{w _1 \times 1000 \times 2}{m _1\left(w _1+w _2\right)}$
Note Volume $=\dfrac{\text { Total mass of solution }}{\text { Density }}=\left(\dfrac{w _1+w _2}{2}\right) mL$
Molality $=\dfrac{\text { Solute }(\text { moles })}{\text { Solvent }(kg)}=\dfrac{w _1 \times 1000}{m _1 \times w _2}$
Given,
molarity $=$ molality
hence,
$\dfrac{2000 w _1}{m _1\left(w _1+w _2\right)}=\dfrac{1000 w _1}{m _1 w _2}$
$\therefore \quad \dfrac{w _2}{w _1+w _2}=\dfrac{1}{2} \Rightarrow w _1=w _2=1$
$\therefore \quad \dfrac{w _1 m _2}{m _1 w _2}=\dfrac{1}{9} \Rightarrow \dfrac{m _1(\text { solute })}{m _2 \text { (solvent) }}=9$
BETA
