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:
We need to calculate the hardness of a water sample in ppm, expressed in terms of CaCO\(_3\) equivalents, given the mass of MgCl\(_2\) in a known mass of water.

Step 2: Key Formula or Approach:
1. Convert the mass of the hardness-causing salt (MgCl\(_2\)) to its equivalent mass of CaCO\(_3\) using the ratio of their molar masses. The molar mass of CaCO\(_3\) is 100 g/mol.
\[ \text{Mass of CaCO}_3 \text{ equiv.} = (\text{Mass of salt}) \times \frac{\text{Molar mass of CaCO}_3}{\text{Molar mass of salt}} \]
2. Calculate the concentration in ppm (parts per million), defined as mg of CaCO\(_3\) equivalent per kg of water.

Step 3: Detailed Explanation:

1. Calculate CaCO\(_3\) equivalent mass:
- Mass of MgCl\(_2\) = 19 mg.
- Molar mass of MgCl\(_2\) = 95 g/mol.
- Molar mass of CaCO\(_3\) = 100 g/mol.
Using the equivalence formula:
\[ \text{Mass of CaCO}_3 \text{ equiv.} = 19 \text{ mg} \times \frac{100}{95} = \frac{1900}{95} \text{ mg} = 20 \text{ mg} \]
This means 19 mg of MgCl\(_2\) creates the same hardness as 20 mg of CaCO\(_3\).

2. Calculate hardness in ppm:
The mass of the water sample is 2 kg.
\[ \text{Hardness (ppm)} = \frac{\text{Mass of CaCO}_3 \text{ equiv. (mg)}}{\text{Mass of water (kg)}} \]
\[ \text{Hardness (ppm)} = \frac{20 \text{ mg}}{2 \text{ kg}} = 10 \text{ mg/kg} \]
Since 1 mg/kg is equivalent to 1 ppm, the hardness is 10 ppm.

Step 4: Final Answer:
The degree of hardness is 10 ppm.