20 mL of 0.1 M solution of compound `NaCO_(3).NaHCO_(3).2H_(2)O` is titrated against 0.05 M HCL. X mL of HCL is used when phenolphthalein is used as an indicator and y mL of HCL is used when methly orange is the indicator in two separate titrations. Hence (y-x) is:

Options

A. 40 mL
B. 80 mL
C. 120 mL
D. None of these

Answer

To solve the problem, we need to find the volumes of HCl used in two separate titrations with different indicators, phenolphthalein and methyl orange, and then calculate the difference \( y - x \).

### Step-by-Step Solution:

1. **Identify the Compound and Its Concentration:**
The compound is sodium carbonate and sodium bicarbonate with two water molecules: \( \text{Na}_2\text{CO}_3 \cdot \text{NaHCO}_3 \cdot 2\text{H}_2\text{O} \). The concentration of the solution is given as 0.1 M and the volume is 20 mL.

2. **Calculate the Number of Milliequivalents of the Compound:**
- The total moles of the compound in the solution can be calculated using the formula:
\[ \text{Moles} = \text{Molarity} \times \text{Volume (L)} \]
\[ \text{Moles} = 0.1 \, \text{mol/L} \times 0.020 \, \text{L} = 0.002 \, \text{mol} \]
- The number of equivalents (N factor for \( \text{Na}_2\text{CO}_3 \) is 2) is:
\[ \text{Milliequivalents} = \text{Moles} \times \text{N factor} \times 1000 \]
\[ = 0.002 \, \text{mol} \times 2 \times 1000 = 4 \, \text{mEq} \]

3. **Titration with Phenolphthalein:**
- When phenolphthalein is used, only \( \text{Na}_2\text{CO}_3 \) reacts with HCl:
\[ \text{Na}_2\text{CO}_3 + 2 \text{HCl} \rightarrow 2 \text{NaCl} + \text{H}_2\text{O} + \text{CO}_2 \]
- Since only half of the \( \text{Na}_2\text{CO}_3 \) is neutralized, the milliequivalents of HCl used (x) will be equal to the milliequivalents of \( \text{Na}_2\text{CO}_3 \):
\[ x = 4 \, \text{mEq} \]
- Using the formula for milliequivalents of HCl:
\[ x = \text{Molarity} \times \text{Volume} \times \text{N factor} \]
\[ 4 = 0.05 \times x \times 1 \]
\[ x = \frac{4}{0.05} = 80 \, \text{mL} \]

4. **Titration with Methyl Orange:**
- When methyl orange is used, both \( \text{Na}_2\text{CO}_3 \) and \( \text{NaHCO}_3 \) react with HCl:
\[ \text{Na}_2\text{CO}_3 + 2 \text{HCl} \rightarrow 2 \text{NaCl} + \text{H}_2\text{O} + \text{CO}_2 \]
\[ \text{NaHCO}_3 + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{CO}_3 \]
- The total milliequivalents of both compounds will be:
\[ \text{Milliequivalents of } \text{Na}_2\text{CO}_3 = 4 \, \text{mEq} \]
\[ \text{Milliequivalents of } \text{NaHCO}_3 = 0.1 \times 20 \times 1 = 2 \, \text{mEq} \]
- Therefore, the total milliequivalents of acid used (y) is:
\[ y = 4 + 2 = 6 \, \text{mEq} \]
- Using the formula for milliequivalents of HCl:
\[ 6 = 0.05 \times y \times 1 \]
\[ y = \frac{6}{0.05} = 120 \, \text{mL} \]

5. **Calculate \( y - x \):**
\[ y - x = 120 - 80 = 40 \, \text{mL} \]

### Final Answer:
Thus, \( y - x = 40 \, \text{mL} \).