1. Given concentration of hardness-causing salt

The water sample contains $10^{-3}\,\text{M } MgSO_4$

This means: $10^{-3}\text{ mol } MgSO_4 \text{ per litre}$

2. Convert }MgSO_4\text{ into }CaCO_3\text{ equivalent}

Hardness is expressed in terms of $CaCO_3$ equivalents.

Using equivalent weights:

• For $MgSO_4$: $\text{Equivalent weight} = \frac{120.37}{2} = 60.185$ because $Mg^{2+}$ has valency $2$.

• For $CaCO_3$: $\text{Equivalent weight} = \frac{100}{2} = 50$

Now, mass of $MgSO_4$ present per litre: $10^{-3} \times 120.37 = 0.12037\,\text{g/L} = 120.37\,\text{mg/L}$

Its $CaCO_3$ equivalent is: $\text{CaCO}_3\text{ equivalent} = 120.37 \times \frac{50}{60.185}$

Since $60.185 = \frac{120.37}{2}$ we get $120.37 \times \frac{50}{60.185} = 120.37 \times \frac{50}{120.37/2} = 100\,\text{mg/L}$

3. Convert to ppm

For dilute aqueous solutions: $1\,\text{mg/L} \approx 1\,\text{ppm}$

Therefore, $100\,\text{mg/L} = 100\,\text{ppm}$

4. Final answer

The hardness expressed as $CaCO_3$ equivalent is: $\boxed{100\,\text{ppm}}$