Question

An aqueous solution of $NaOH$ having density $1.1$ $kg/{dm}^3$ contains $0.02$ mole fraction of $NaOH$. The $\%$ by mass of $NaOH$ in the solution is :

• A. $4.34$
• B. $2.17$
• C. $5.28$
• D. $8.34$

Answer 1

The correct option is A $4.34$

The mole fraction of NaOH is 0.02 mole.
Let,

Mass of water in the solution = 1000g

1000 g of water corresponds to 55.55 moles of water.

Mole fraction of solute $=\frac{n}{55.55+n}=0.02=\frac{1}{50}$

$\Rightarrow 50n=55.55+n$

$\Rightarrow 49n=55.55$

$\Rightarrow n=\frac{55.55}{49}=1.13$ mol

Mass of NaOH $=1.13\times 40=45.3$ g.

$\therefore$ % by mass of NaOH $=\frac{\text{Mass of NaOH}}{\text{Mass of NaOH}+\text{Mass of water}}\times 100$

$=\frac{45.3}{45.3+1000}=4.34$ %.

Option A is correct.