Question

The density of a $3\text{M}$ $N{a}_{2}C{O}_3$ solution in water is $\text{1.2 g/L}$. The percentage by weight and ppm of $N{a}_{2}C{O}_3$ are respectively:

• A. $26.5$ and $2.65\times {10}^5$
• B. $2.65$ and $2.65\times {10}^4$
• C. 265 and $2.65\times {10}^3$
• D. 5.23 and $2.65\times {10}^6$

Answer 1

The correct option is A $26.5$ and $2.65\times {10}^5$

Let the volume of the solution be $1000\text{mL}$
$\Rightarrow \text{Density of solution}=\frac{\text{mass of solution}}{\text{volume of solution}}\Rightarrow 1.2=\frac{\text{mass of solution}}{1000}$
Mass of solution = $1200\text{g}$
$\Rightarrow \text{Molarity}=\frac{\text{moles of solute of}N{a}_{2}C{O}_3}{\text{total volume of solution in mL}}\times 1000$
$\Rightarrow 3M=\frac{\text{moles of solute}}{1000}\times 1000$
Moles of $N{a}_{2}C{O}_3=3$ moles
So, mass of $N{a}_{2}C{O}_3$ in solution $=3\text{mol}\times 106\text{g/mol}=318\text{g}$
$\%\frac{w}{w}=\frac{\text{weight of}N{a}_{2}C{O}_3\text{in solution (g)}}{\text{total weight of the solution (g)}}\times 100$
$\%\frac{w}{w}=\frac{318}{1200}\times 100=26.5\%$

$\text{ppm of}N{a}_{2}C{O}_3=\frac{\text{weight of component (g)}}{\text{total weight of solution (g)}}\times {10}^6$
$=\frac{318}{1200}\times {10}^6=2.65\times {10}^5$