Question:
A 2 kg water sample contains 408 mg of $CaSO_{4}$ (M.wt=136). The hardness in terms of $CaCO_{3}$ equivalents (in ppm) is
A 2 kg water sample contains 408 mg of $CaSO_{4}$ (M.wt=136). The hardness in terms of $CaCO_{3}$ equivalents (in ppm) is
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Hardness (ppm) is the mg of $CaCO_{3}$ equivalent per 1 kg of water. Always divide by the sample weight if it's not 1 kg.
Updated On: Jul 4, 2026
- 100
- 136
- 150
- 204
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The Correct Option is C
Solution and Explanation
Step 1: Concept
Hardness is expressed as $CaCO_{3}$ equivalent $= \text{Mass of salt} \times \frac{\text{Eq. wt of } CaCO_{3}}{\text{Eq. wt of salt}}$.
Step 2: Meaning
$CaSO_{4}$ (M.wt 136) has valency 2, so Eq. wt $= 68$. $CaCO_{3}$ (M.wt 100) has valency 2, so Eq. wt $= 50$.
Step 3: Analysis
$CaCO_{3}$ equivalent $= 408 \text{ mg} \times (50/68) = 300$ mg in 2 kg of water.
Step 4: Conclusion
Hardness in ppm (mg/L or mg/kg) $= 300 \text{ mg} / 2 \text{ kg} = 150$ ppm. Final Answer: (C)
Hardness is expressed as $CaCO_{3}$ equivalent $= \text{Mass of salt} \times \frac{\text{Eq. wt of } CaCO_{3}}{\text{Eq. wt of salt}}$.
Step 2: Meaning
$CaSO_{4}$ (M.wt 136) has valency 2, so Eq. wt $= 68$. $CaCO_{3}$ (M.wt 100) has valency 2, so Eq. wt $= 50$.
Step 3: Analysis
$CaCO_{3}$ equivalent $= 408 \text{ mg} \times (50/68) = 300$ mg in 2 kg of water.
Step 4: Conclusion
Hardness in ppm (mg/L or mg/kg) $= 300 \text{ mg} / 2 \text{ kg} = 150$ ppm. Final Answer: (C)
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