Question

If $x$ g of $N{a}_{2}C{O}_3$ of $95\%$ purity is required to neutralise $45.6$ mL of $0.235N$ acid, then the value of $10000x$ is:

• A. $5978$
• B. $6000$
• C. $5500$
• D. none of the above

Answer 1

The correct option is A $5978$

Milliequivalents of $N{a}_{2}C{O}_3=$ Milliequivalents of acid
$\frac{\text{Weight}\times 1000}{\text{Ew}}=V$ (in mL) $\times N$

$\frac{W_2\times 1000}{53}=45.6\times 0.235$ $⟹W_2=0.5679$ g

Molecular weight of $N{a}_{2}C{O}_3=23\times 2+12+16\times 3=106$ g/mol
$\text{Ew}=\frac{\text{Mw}}{2}=\frac{106}{2}=53$ g
Weight of $N{a}_{2}C{O}_3$ of $95\%$ purity $=\frac{0.5679\times 100}{95}$ $=0.5978$ g