One litre of a sample of hard water contains 1 mg of CaCl₂ and 1 mg of MgSO₄. What is the total hardness in terms of parts of CaCO₃ per million parts of water by mass?

1.95 ppm

Let's start with the given information:

\[\begin{align*}1 \ \text{mole of CaCl}_2 &\equiv 1 \ \text{mole of CaCO}_3 \\ 1 \ \text{mole of MgCl}_2 &\equiv 1 \ \text{mole of CaCO}_3 \\ \end{align*} \]

From this, we can find the conversion factors:

\[\begin{align*} 100 \ \text{mg CaCO}_3 &\text{ is produced by } 95 \ \text{mg MgCl}_2 \\ 1 \ \text{mg of MgCl}_2 &\text{ gives } \frac{95}{100} \ \text{mg CaCO}_3 = 1.05 \ \text{mg CaCO}_3 \\ 1 \ \text{mg of CaCl}_2 &\text{ gives } \frac{111}{100} \ \text{mg CaCO}_3 = 0.90 \ \text{mg CaCO}_3 \\ \end{align*}\]

Now, let's find the total \(CaCO_3\) per liter of water:

\[\begin{align*} \text{Total CaCO}_3 \ \text{per liter of water} &= 1.05 \ \text{mg} + 0.90 \ \text{mg} = 1.95 \ \text{mg} \\
\end{align*}\]

The weight of 1000 ml (1 liter) of water is  \(10^6\) mg.

\[\begin{align*}\text{Total hardness in terms of parts of CaCO}_3 \ \text{per} \ 10^6 \ \text{parts of water by weight} &= \frac{1.95 \ \text{mg}}{10^6 \ \text{mg}} \times 10^6 \\&= 1.95 \ \text{ppm}
\end{align*}\]

Therefore, the total hardness is \(1.95 \ \text{ppm}\).