Question

8g of $NaOH$ is dissloved in 18g of $H_{2}O$. Mole fraction of $NaOH$ in solution and molality $\left(inmolk{g}^{-1}\right)$ of the solutions respectively are :

• A. $0.167,11.11$
• B. $0.2,22.20$
• C. $0.2,11.11$
• D. $0.167,22.20$

Answer 1

The correct option is A $0.167,11.11$

8 g NaOH
$\text{mol of}NaOH=\frac{8}{40}=0.2mol$
18 g $H_{2}O$
$\text{mol of}{H}_{2}O=\frac{18}{18}=1mol$
Mole fraction
$X_{NaOH}=\frac{0.2}{1.2}=0.167$
$\text{Molality}=\frac{0.2\times 1000}{18}=11.11molk{g}^{-1}$
Option A is the correct answer