Question

$27.2$ mg of $CaSO_4$ and $2.4$ mg of $MgSO_4$ are present in a $2$ kg water sample. What is the total hardness of water (in ppm) in terms of equivalents of $CaCO_3$? (molecular weight of $CaSO_4 = 136$ & molecular weight of $MgSO_4 = 120$)

• A. $11$
• B. $10$
• C. $20$
• D. $22$

Hint:

The molecular weight of $CaCO_3$ is exactly $100$, which makes it the standard reference for hardness. Always remember to divide the final mass by the volume of the water sample if it is not $1$ Litre.

Answer 1

The Correct Option is A

1. Calculate Hardness due to $CaSO_4$: Hardness in terms of $CaCO_3$ equivalent = $\text{Mass of salt} \times \frac{\text{Mol. wt. of } CaCO_3}{\text{Mol. wt. of salt}}$ $$\text{Equivalent of } CaSO_4 = 27.2 \text{ mg} \times \frac{100}{136} = 0.2 \times 100 = 20 \text{ mg}\lt strong\gt 2. Calculate Hardness due to $MgSO_4$:\lt /strong\gt \text{Equivalent of } MgSO_4 = 2.4 \text{ mg} \times \frac{100}{120} = 0.02 \times 100 = 2 \text{ mg}\lt strong\gt 3. Total $CaCO_3$ equivalent:\lt /strong\gt \text{Total equivalent} = 20 \text{ mg} + 2 \text{ mg} = 22 \text{ mg}$$

4. Convert to ppm (mg/L): The sample size is $2$ kg, which is approximately $2$ Litres. Hardness in ppm is the amount of $CaCO_3$ equivalent in $1$ Litre of water. $$\text{Total Hardness (ppm)} = \frac{22 \text{ mg}}{2 \text{ L}} = 11 \text{ mg/L or ppm}$$ Therefore, the total hardness of the water sample is $11$ ppm.