Question

What is the degree of hardness (in ppm) of a sample containing 19 mg of MgCl\(_2\) (Molecular Weight = 95) in 2 kg water sample? (express it in terms of equivalents of CaCO\(_3\))

• A. 10
• B. 20
• C. 30
• D. 40

Hint:

Hardness calculations always involve converting the given salt into its CaCO\(_3\) equivalent.
The conversion factor is always \(\frac{100}{\text{Molar mass of the salt}}\).
Then, remember that ppm = mg/kg (or mg/L for dilute solutions).

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The objective is to calculate the hardness of water expressed as CaCO₃ equivalent in ppm.

Step 2: Key Formula or Approach:
Find the CaCO₃ equivalent of MgCl₂ and express the concentration as mg of CaCO₃ per kg of water.

Step 3: Detailed Explanation:
The sample contains 19 mg of MgCl₂ dissolved in 2 kg of water. Its CaCO₃ equivalent is \(19\times100/95=20\) mg. Dividing this equivalent mass by 2 kg gives \(20/2=10\) mg/kg. Since mg/kg and ppm are numerically identical for water, the hardness equals 10 ppm.

Step 4: Final Answer:
The calculated hardness is 10 ppm.