A 50.0 mL sample of hard water is acidified with dilute hydrochloric acid and boiled…

Step 1. $n(\text{BaSO}_4) = \dfrac{0.466\ \text{g}}{233\ \text{g/mol}} = 2.00\ \text{mmol}$. This is the amount of sulphate ions in the sample.

Step 2. $c = \dfrac{2.00\ \text{mmol}}{50.0\ \text{mL}} = 0.0400\ \text{mol/L}$.

Step 3. $M(\text{SO}_4^{2-}) = 96\ \text{g/mol}$. The mass for each litre is $0.0400\ \text{mol/L} \times 96\ \text{g/mol} = 3.84\ \text{g/L}$.