Question

$25\text{mL}$ of a solution of $2\text{m}$ $N{a}_{2}C{O}_3$ having a specific gravity of $1.212\text{g/mL}$ requires what volume of a solution of $HCl$, containing $18.25\text{g}$ of the acid per litre, for complete neutralization?

• A. $0.05\text{L}$
• B. $0.2\text{L}$
• C. $0.5\text{L}$
• D. $1\text{L}$

Answer 1

The correct option is B $0.2\text{L}$

For the neutralization reaction,
$N{a}_{2}C{O}_3+2HCl\to 2NaCl+H_{2}O+C{O}_2$
Given,
Specific gravity for $N{a}_{2}C{O}_3$ solution is $1.212\text{g/mL}$.
Let us assume that mass of the solvent be $1000\text{g}$.
Thus, moles of sodium carbonate $=2\text{mol}$
Again, molar mass of $N{a}_{2}C{O}_3=23\times 2+12+16\times 3=106\text{g/mol}$
mass of sodium carbonate will be $=2\times 106\text{g}=212\text{g}$

$\therefore$ mass of the solvent + solute
$=1000+212=1212\text{g}$

Hence volume of the solution $=\frac{\text{mass}}{\text{density}}=\frac{1212}{1.212}=1000\text{mL}$

molarity of sodium carbonate $=\frac{\text{moles}}{\text{volume}}=\frac{2}{1}=2\text{M}$

We know,
n-factor of sodium carbonate $=2$
n-factor of $HCl=1$
$\text{n-factor}\times \text{molarity}\times \text{volume of the base}=\text{n-factor}\times \text{number of moles of the acid}$
$2\times 2\times 0.025=1\times \text{number of moles of}HCl$

Hence, moles of $HCl=0.1$
mass of $HCl=\text{moles}\times \text{molar mass}$
$=0.1\times 36.5=3.65\text{g}$
Given, $1$ litre of the $HCl$ solution contains $18.25\text{g}$ of $HCl$
volume of the solution needed $=\frac{1}{18.25}\times 3.65=0.2\text{L}$