Question

25 mL of a solution of $N{a}_{2}C{O}_3$ having a specific gravity of 1.25 g/mL requires what volume of a solution of $HCl$, containing 215.3 g of the acid per litre, for complete neutralization?

• A. 0.05 L
• B. 0.1 L
• C. 0.2 L
• D. 0.25 L

Answer 1

The correct option is B 0.1 L

For the neutralization reaction,
$N{a}_{2}C{O}_3+2HCl\to 2NaCl+H_{2}O+C{O}_2$
Since specific gravity for $N{a}_{2}C{O}_3$ is 1.25 g/mL
So, mass of $N{a}_{2}C{O}_3$ = $1.25\times 25=31.25g$
$\therefore$ moles of $N{a}_{2}C{O}_3$ = $\frac{mass}{molarmass}=\frac{31.25}{106}=0.295$ mol
Now, from stoichiometry
1 mol $N{a}_{2}C{O}_3$ reacts with 2 moles of $HCl$
So, 0.295 moles $N{a}_{2}C{O}_3$ react with 0.295$\times$2 moles of $HCl$
Mass of $HCl$ = $moles\times molarmass=0.295\times 2\times 36.5=21.53g$
$\therefore$ volume of $HCl$ solution = $\frac{21.53g}{215.3g/L}=0.1L$