`0.7 g "of" Na_(2)CO_(3).xH_(2)O` were dissolved in water and the volume was made to `100 mL, 20 mL` of this solution required `19.8 mL "of" N//10 HCl` for complete neutralization. The value of `x` is:

Options

A. `7`
B. `3`
C. `2`
D. `5`

Explanation

To find the value of \( x \) in the compound \( \text{Na}_2\text{CO}_3 \cdot x\text{H}_2\text{O} \), we can follow these steps:

### Step 1: Calculate the amount of sodium carbonate in 1000 mL
Given that 0.7 g of \( \text{Na}_2\text{CO}_3 \cdot x\text{H}_2\text{O} \) is dissolved in 100 mL, we can find the amount in 1000 mL:
\[ \text{Amount in 1000 mL} = \frac{0.7 \, \text{g}}{100 \, \text{mL}} \times 1000 \, \text{mL} = 7 \, \text{g} \]

### Step 2: Determine the normality of the sodium carbonate solution
We know that 20 mL of the sodium carbonate solution neutralizes 19.8 mL of \( \frac{N}{10} \) HCl. The normality of the sodium carbonate solution can be calculated using the formula:
\[ N_1V_1 = N_2V_2 \]
Where:
- \( N_1 \) = Normality of \( \text{Na}_2\text{CO}_3 \)
- \( V_1 \) = Volume of \( \text{Na}_2\text{CO}_3 \) solution = 20 mL
- \( N_2 \) = Normality of HCl = \( \frac{1}{10} \)
- \( V_2 \) = Volume of HCl = 19.8 mL

Substituting the values:
\[ N_1 \times 20 = \frac{1}{10} \times 19.8 \]
\[ N_1 = \frac{19.8}{200} = 0.099 \, N \]

### Step 3: Relate normality to strength and equivalent weight
Normality is also defined as:
\[ N = \frac{\text{Strength (g/L)}}{\text{Equivalent weight (g/equiv)}} \]
The strength of the sodium carbonate solution in 1000 mL is 7 g. Thus:
\[ 0.099 = \frac{7}{\text{Equivalent weight of } \text{Na}_2\text{CO}_3} \]
From this, we can calculate the equivalent weight:
\[ \text{Equivalent weight of } \text{Na}_2\text{CO}_3 = \frac{7}{0.099} \approx 70.70 \, \text{g/equiv} \]

### Step 4: Calculate the molar mass of sodium carbonate
The equivalent weight is related to the molar mass by the formula:
\[ \text{Equivalent weight} = \frac{\text{Molar mass}}{n} \]
Where \( n \) is the number of equivalents. For \( \text{Na}_2\text{CO}_3 \), \( n = 2 \) because it can donate 2 moles of \( \text{OH}^- \) ions.

### Step 5: Set up the equation for molar mass
Let’s calculate the molar mass of \( \text{Na}_2\text{CO}_3 \cdot x\text{H}_2\text{O} \):
\[ \text{Molar mass} = 23 \times 2 + 12 + 16 \times 3 + 18x = 46 + 12 + 48 + 18x = 106 + 18x \]
Setting the equivalent weight equal to the calculated value:
\[ \frac{106 + 18x}{2} = 70.70 \]
Multiplying both sides by 2:
\[ 106 + 18x = 141.4 \]
Solving for \( x \):
\[ 18x = 141.4 - 106 \]
\[ 18x = 35.4 \]
\[ x = \frac{35.4}{18} \approx 1.9667 \approx 2 \]

### Final Answer
Thus, the value of \( x \) is approximately \( 2 \).

---