Question

The density of a 3 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 mL
$\Rightarrow Densityofsolution=\frac{\text{mass of solution}}{\text{volume of solution}}1.2=\frac{\text{mass of solution}}{1000}$
Mass of solution = 1200 g
$\Rightarrow Molarity=\frac{\text{moles of solute of}N{a}_{2}C{O}_3}{\text{total volume of solution in mL}}\times 1000$
$3M=\frac{\text{moles of solute}}{1000}\times 1000$

Moles of $N{a}_{2}C{O}_3=3$ moles
So, mass of $N{a}_{2}C{O}_3$ in solution $=3mol\times$ 106 g/mol = 318 g
% $\frac{w}{w}=\frac{\text{weight of}N{a}_{2}C{O}_3\text{in solution (g)}}{\text{total weight of the solution (g)}}\times 100$
% $\frac{w}{w}=\frac{318}{1200}\times 100=26.5\%$
ppm of $N{a}_{2}C{O}_3=\frac{\text{weight of component (g)}}{\text{total weight of solution (g)}}\times {10}^6$
ppm of $N{a}_{2}C{O}_3=\frac{318}{1200}\times {10}^6=2.65\times {10}^5$