Question

The mass of ${Na}_{2}C{O}_3$ of $95\%$ purity that would be required to neutralise $45.5$ mL of $0.235$ N acid is:

(write in multiples of 10 (only the nearest single digit))

Answer 1

$\because$ Meq. of ${Na}_{2}C{O}_3$ $=$ Meq. of $H_{2}S{O}_4$, 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
If $95$ g pure ${Na}_{2}C{O}_3$ is to be taken then mass of sample $=$ $100$ g
$\therefore$ $0.5679$ g pure ${Na}_{2}C{O}_3$ would require mass of sample $=$ $100\times \frac{0.5679}{95}=0.5978$ g
So, the answer is $6$.