What is the degree of hardness of a sample of water containing 24 mg of $MgS{O_4}$ (molecular mass 120) per kg of water?

A. 10 ppm

B. 15 ppm

C. 20 ppm

D. 25 ppm

To determine the degree of hardness of the water sample containing magnesium sulfate ($MgSO_4$), we need to calculate the concentration of the dissolved substance in parts per million (ppm). The formula for ppm is straightforward: it represents the mass of solute per million parts of solution, typically expressed as milligrams of solute per liter of solution. In this case, we have 24 mg of $MgSO_4$ per kg of water, which is equivalent to 1 liter of water since the density of water is approximately 1 kg/L.

Calculating Hardness in ppm

Given that we have 24 mg of $MgSO_4$ in 1 kg (or 1 L) of water, we can directly express this in ppm:

• Mass of $MgSO_4$ = 24 mg
• Volume of water = 1 L = 1000 g

To find the hardness in ppm, we can use the following relationship:

ppm = (mass of solute in mg / volume of solution in L)

Substituting the values we have:

ppm = (24 mg / 1 L) = 24 ppm

Understanding the Options

Looking at the options provided:

• A. 10 ppm
• B. 15 ppm
• C. 20 ppm
• D. 25 ppm

Since our calculation shows that the hardness is 24 ppm, it does not exactly match any of the options. However, the closest option is D, which is 25 ppm. This slight discrepancy could arise from rounding or measurement variations in practical scenarios.

Conclusion

In summary, the degree of hardness of the water sample containing 24 mg of $MgSO_4$ per kg of water is approximately 24 ppm, making option D (25 ppm) the most appropriate choice among the given answers. Understanding how to convert mass to concentration in ppm is crucial in various applications, including environmental science and water quality assessment.