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. **Understanding the Reaction**:
The reaction between sodium carbonate (\( Na_2CO_3 \)) and sulfuric acid (\( H_2SO_4 \)) can be represented as:
\[ Na_2CO_3 + H_2SO_4 \rightarrow Na_2SO_4 + CO_2 + H_2O \]
From this reaction, we see that 1 mole of \( Na_2CO_3 \) reacts with 1 mole of \( H_2SO_4 \).

2. **Normality of \( H_2SO_4 \)**:
The given solution of sulfuric acid is \( N/10 \). This means:
\[ \text{Normality of } H_2SO_4 = 0.1 \, N \]
Since \( H_2SO_4 \) is a diprotic acid, 1 mole of \( H_2SO_4 \) provides 2 equivalents of \( H^+ \). Therefore, \( N/10 \) means that 1 L of this solution contains 0.1 equivalents of \( H^+ \).

3. **Calculating the Number of Equivalents of \( H_2SO_4 \)**:
For 100 ml (0.1 L) of \( N/10 \) \( H_2SO_4 \):
\[ \text{Number of equivalents} = \text{Normality} \times \text{Volume in L} = 0.1 \, N \times 0.1 \, L = 0.01 \, \text{equivalents} \]

4. **Finding the Equivalent Weight of \( Na_2CO_3 \cdot xH_2O \)**:
The equivalent weight of \( Na_2CO_3 \) is calculated as:
\[ \text{Molar mass of } Na_2CO_3 = 2 \times 23 + 12 + 3 \times 16 = 106 \, g/mol \]
Since it can donate 1 equivalent of \( Na^+ \) (or react with 1 equivalent of acid), the equivalent weight is:
\[ \text{Equivalent weight} = 106 \, g/mol \]

5. **Setting Up the Equation**:
The weight of \( Na_2CO_3 \cdot xH_2O \) is given as 0.62 g. The number of equivalents of \( Na_2CO_3 \cdot xH_2O \) can be expressed as:
\[ \text{Number of equivalents} = \frac{\text{Weight}}{\text{Equivalent weight}} = \frac{0.62 \, g}{106 \, g/equiv} = 0.00585 \, \text{equivalents} \]

6. **Equating the Equivalents**:
Since the number of equivalents of \( Na_2CO_3 \cdot xH_2O \) must equal the number of equivalents of \( H_2SO_4 \):
\[ 0.00585 \, \text{equivalents} = 0.01 \, \text{equivalents} \]
This indicates that the weight of the hydrated salt must be adjusted for the water of hydration.

7. **Calculating the Value of \( x \)**:
The molar mass of \( Na_2CO_3 \cdot xH_2O \) can be expressed as:
\[ \text{Molar mass} = 106 + 18x \]
The weight of \( Na_2CO_3 \cdot xH_2O \) can be calculated as:
\[ 0.62 = \text{Number of moles} \times (106 + 18x) \]
Since we have 0.00585 equivalents, we can find the moles:
\[ \text{Moles} = \frac{0.00585}{1} = 0.00585 \]
Thus:
\[ 0.62 = 0.00585 \times (106 + 18x) \]
Solving for \( x \):
\[ 106 + 18x = \frac{0.62}{0.00585} \approx 106.67 \]
\[ 18x = 106.67 - 106 \]
\[ 18x = 0.67 \implies x \approx \frac{0.67}{18} \approx 0.037 \]

8. **Final Value of \( x \)**:
Since \( x \) must be a whole number, we can conclude that \( x = 0 \) as it indicates that there is no water of hydration.

### Conclusion:
The value of \( x \) in \( Na_2CO_3 \cdot xH_2O \) must be 0.