Permanent hardness is due to `CI^(ɵ)` and `SO_4^(2-)` of `Mg^(2+)` and `Ca^(2+)` and is removed by adding `Na_2CO_3`.
`{:(CaSO_(4)+Na_(2)CO_(3)toCaCO_(3)+Na_(2)SO_(4)),(CaCl_(2)+Na_(2)CO_(3)toCaCO_(3)+2NaCl):}` Which of the following statements is`//`are correct?

Options

A. If hardness is `100 p p m CaCO_(3)` the amount of `Na_(2)CO_(3)` required to soften `10 L` of hard water is `10.6 g`
B. If hardness is `100 p p m CaCO_(3)`, the amount of `Na_(2)CO_(3)` required to soften `10L` of hard water is `1.06 g`.
C. If hardness is `420 p p m MgCO_(3)`, the amount of `Na_(2)CO_(3)`required to soften `10 L` of hard water is `53 g`
D. If hardness is `420 p p m MgCO_(3)` is the amount of `Na_(2)CO_(3)` required to soften `10 L` of hard water is `5.3 g`

Explanation

To solve the problem regarding the removal of permanent hardness in water caused by the presence of chloride and sulfate salts of magnesium and calcium, we will follow a step-by-step approach.

### Step 1: Understanding Permanent Hardness
Permanent hardness in water is primarily due to the presence of calcium and magnesium salts, specifically CaCl₂, CaSO₄, MgCl₂, and MgSO₄. These salts do not precipitate upon boiling and thus contribute to the hardness of water.

### Step 2: Reaction with Sodium Carbonate
To remove this hardness, sodium carbonate (Na₂CO₃) is added. The relevant reactions are:
1. For calcium sulfate:
\[ \text{CaSO}_4 + \text{Na}_2\text{CO}_3 \rightarrow \text{CaCO}_3 \downarrow + \text{Na}_2\text{SO}_4 \]
Here, calcium carbonate (CaCO₃) precipitates out, which can be filtered out.

2. For calcium chloride:
\[ \text{CaCl}_2 + \text{Na}_2\text{CO}_3 \rightarrow \text{CaCO}_3 \downarrow + 2\text{NaCl} \]
Again, calcium carbonate precipitates out.

### Step 3: Calculating the Amount of Na₂CO₃ Required
Given that the hardness of water is 100 parts per million (ppm) of CaCO₃, we need to calculate how much Na₂CO₃ is required to soften 10 liters of this water.

1. **Convert ppm to grams**:
- 100 ppm means 100 grams of CaCO₃ in 1,000,000 grams of water.
- Since the density of water is approximately 1 g/mL, 10 liters of water weighs about 10,000 grams.

2. **Calculate the mass of CaCO₃ in 10 liters**:
\[ \text{Mass of CaCO}_3 = \frac{100 \text{ g}}{1,000,000 \text{ g}} \times 10,000 \text{ g} = 1 \text{ g} \]

3. **Calculate moles of CaCO₃**:
- Molar mass of CaCO₃ = 100 g/mol.
\[ \text{Moles of CaCO}_3 = \frac{1 \text{ g}}{100 \text{ g/mol}} = 0.01 \text{ moles} \]

4. **Stoichiometry of the reaction**:
- From the reaction, 1 mole of Na₂CO₃ is required for 1 mole of CaCO₃.
- Therefore, moles of Na₂CO₃ required = 0.01 moles.

5. **Calculate the mass of Na₂CO₃ required**:
- Molar mass of Na₂CO₃ = 106 g/mol.
\[ \text{Mass of Na}_2\text{CO}_3 = 0.01 \text{ moles} \times 106 \text{ g/mol} = 1.06 \text{ g} \]

### Conclusion
The amount of Na₂CO₃ required to soften 10 liters of water with a hardness of 100 ppm CaCO₃ is **1.06 grams**.