Question

Calculate the amount of 95% pure $N{a}_{2}C{O}_3$ required to prepare 5 litre of 0.5 M solution.

Answer 1

$1L$ of $0.5M$ contains $0.5$ moles

$5L$ of $0.5M$ contains $0.5\times 5=2.5$ moles

Molar mass of $N{a}_{2}C{O}_3=(23\times 2)+12+(16\times 3)=106g$

$\Rightarrow 2.5\times 106=265g$ for a $100$% solution.

For $95$% pure $N{a}_{2}C{O}_3$,

$\frac{265}{0.95}=278.98g$ of $N{a}_{2}C{O}_3$ is required.