Question

The number of grams of anhydrous $N{a}_{2}C{O}_3$ present in 250 mL of 0.25 M solution is :

• A. 6.625 g
• B. 0.625 g
• C. 0.567 g
• D. 7.125 g

Answer 1

The correct option is A 6.625 g

Molarity (M) $=$ 0.25 M

Volume (V) $=$ 250 ml

Molecular weight of $N{a}_{2}C{O}_3$ (M) $=(23\times 2)+12+(16\times 3)=106$

$Molarity\left(M\right)=\frac{m\times 1000}{M\times V\left(ml\right)}$

$0.25=\frac{m\times 1000}{106\times 250}$

$m=\frac{25\times 106}{4}$

$m=0.25\times 26.5$

$m=6.625g$