Question

The mass of $N{a}_{2}C{O}_3$ required to prepare $500$ ml of $0.1$ M solution is:

• A. 10.6 g
• B. 5.3 g
• C. 2.65 g
• D. 7.95 g

Answer 1

The correct option is B 5.3 g

Weight of $N{a}_{2}C{O}_3=\frac{Molarity\times Mo1.wt.\times V}{1000}$ (where $V$ is in ml).

Substituting values in the above equation, we get-

Weight of $N{a}_{2}C{O}_3=\frac{0.1\times 106\times 500}{1000}=5.3$ g.