0.62 g `Na_2 CO_3. xH_2 O` completely neutralises 100 ml of N/10 `H_2 SO_4`. The value of x must be:

Options

A. 1
B. 6
C. 8
D. 10

Answer

To solve the problem, we need to determine the value of \( x \) in the compound \( Na_2CO_3 \cdot xH_2O \) that completely neutralizes 100 ml of \( N/10 \) \( H_2SO_4 \).

### Step-by-Step Solution:

1. **Write the Neutralization Reaction**:
The reaction between sodium carbonate and sulfuric acid can be represented as:
\[ Na_2CO_3 \cdot xH_2O + H_2SO_4 \rightarrow Na_2SO_4 + CO_2 + xH_2O \]

2. **Calculate Moles of \( H_2SO_4 \)**:
Given that the normality of \( H_2SO_4 \) is \( N/10 \) (which is 0.1 N), and the volume is 100 ml (or 0.1 L), we can calculate the moles of \( H_2SO_4 \):
\[ \text{Normality} = \frac{\text{Moles of solute}}{\text{Volume of solution in L}} \times \text{Basicity} \]
For \( H_2SO_4 \), the basicity is 2 (since it can donate 2 protons). Thus:
\[ \text{Moles of } H_2SO_4 = \text{Normality} \times \text{Volume in L} \times \text{Basicity} = 0.1 \times 0.1 \times \frac{1}{2} = 0.005 \text{ moles} \]

3. **Determine Moles of \( Na_2CO_3 \cdot xH_2O \)**:
From the balanced equation, we see that 1 mole of \( Na_2CO_3 \cdot xH_2O \) reacts with 1 mole of \( H_2SO_4 \). Therefore, the moles of \( Na_2CO_3 \cdot xH_2O \) required will also be 0.005 moles.

4. **Calculate the Molecular Weight of \( Na_2CO_3 \cdot xH_2O \)**:
The weight of \( Na_2CO_3 \cdot xH_2O \) is given as 0.62 g. Using the formula:
\[ \text{Moles} = \frac{\text{Weight}}{\text{Molecular Weight}} \]
We can rearrange this to find the molecular weight:
\[ \text{Molecular Weight} = \frac{0.62 \text{ g}}{0.005 \text{ moles}} = 124 \text{ g/mol} \]

5. **Calculate the Molecular Weight of \( Na_2CO_3 \cdot xH_2O \)**:
The molecular weight of \( Na_2CO_3 \) can be calculated as follows:
- Sodium (Na): 23 g/mol × 2 = 46 g/mol
- Carbon (C): 12 g/mol × 1 = 12 g/mol
- Oxygen (O): 16 g/mol × 3 = 48 g/mol
- Water (H2O): 18 g/mol × x = 18x g/mol

Therefore, the total molecular weight is:
\[ 46 + 12 + 48 + 18x = 124 \]

6. **Solve for \( x \)**:
Rearranging the equation:
\[ 106 + 18x = 124 \]
\[ 18x = 124 - 106 \]
\[ 18x = 18 \]
\[ x = 1 \]

### Conclusion:
The value of \( x \) must be **1**.