Question

The mass of ${Na}_{2}C{O}_3$ of $95\%$ purity that would be required to neutralise $45.5$ mL of $0.235N$ acid is (in multiples of $10$ and to the nearest single digit)

Answer 1

Milliequivalents of ${Na}_{2}C{O}_3=$ Milliequivalents of $H_{2}S{O}_4$ used for neutralization

$\therefore \left(\frac{w}{53}\right)\times 1000=45.6\times 0.235$

$\therefore W_{{Na}_{2}C{O}_3}=0.5679$ g

$\because 95$ g pure${Na}_{2}C{O}_3$ is to be taken, mass of sample $=100$ g

$\therefore 0.5679$ g pure ${Na}_{2}C{O}_3$ is to be taken, mass of sample $=\frac{(100\times 0.5679)}{95}=0.5978$ g

So, the answer is $6$.