Question

$1.25\, g$ of a sample of $Na _{2} CO _{3}$ and $Na _{2} SO _{4}$ is dissolved in $250\, ml$ solution. $25 ml$ of this solution neutralises $20\, ml$ of $0.1\, N\, H _{2} SO _{4}$. The $\%$ of $Na _{2} CO _{3}$ in this sample is

• A. 84.80%
• B. 8.48%
• C. 15.20%
• D. 42.40%

Answer 1

Answer: (A)

84.80%

Answer 2

The correct option is (A): 84.80 %
Let the amount of $Na _{2} CO _{3}$ present in the mixture be $x g$
$Na _{2} SO _{4}$ will not react with $H _{2} SO _{4}$.
Then $\frac{x}{53}=\frac{20 \times 0.1 \times 10}{1000}$
$\therefore x=1.06 \,g$
$\therefore$ Percentage of $Na _{2} CO _{3}=\frac{1.06 \times 100}{1.25}$
$=84.8 \%$