0.848 g aqueous solution of a mixture containing `Na_(2)CO_(3)` NaOH, and an inert matter is titrated with `(M)/(2)` HCl. The colour of phenolphthalein disappears when 20 " mL of " the acid has been added. Methyl orange is then added and 8.0 mL more of the acid is requried to give a red colour to the solution. The percentage of `Na_(2)CO_(3)` is

Options

A. 25
B. 12.5
C. 75
D. 50

Explanation

To solve the problem, we need to determine the percentage of sodium carbonate (Na₂CO₃) in a mixture that also contains sodium hydroxide (NaOH) and inert matter. We will use the information provided about the titration with hydrochloric acid (HCl) and the indicators used (phenolphthalein and methyl orange).

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Total mass of the solution = 0.848 g
- Volume of HCl added until phenolphthalein changes color = 20 mL
- Volume of HCl added after adding methyl orange = 8 mL
- Total volume of HCl = 20 mL + 8 mL = 28 mL
- Concentration of HCl = M/2 = 0.5 M

2. **Calculate the Milliequivalents of HCl:**
- Milliequivalents of HCl used = Concentration (M) × Volume (mL)
- For the first part (20 mL):
\[ \text{Milliequivalents of HCl} = 0.5 \, \text{mol/L} \times 20 \, \text{mL} = 0.01 \, \text{mol} = 10 \, \text{meq} \]
- For the second part (8 mL):
\[ \text{Milliequivalents of HCl} = 0.5 \, \text{mol/L} \times 8 \, \text{mL} = 0.004 \, \text{mol} = 4 \, \text{meq} \]

3. **Set Up the Reactions:**
- The first reaction (with phenolphthalein):
\[ \text{Na}_2\text{CO}_3 + \text{HCl} \rightarrow \text{NaHCO}_3 + \text{NaCl} \]
- The second reaction (with methyl orange):
\[ \text{NaHCO}_3 + \text{HCl} \rightarrow \text{H}_2\text{CO}_3 + \text{NaCl} \]

4. **Calculate the Milliequivalents of Na₂CO₃ and NaOH:**
- Let the weight of Na₂CO₃ be \( x \) g and the weight of NaOH be \( y \) g.
- The milliequivalents of Na₂CO₃ reacting with HCl:
\[ \text{Milliequivalents of Na}_2\text{CO}_3 = \frac{x}{106} \times 1000 \]
- The milliequivalents of NaOH reacting with HCl:
\[ \text{Milliequivalents of NaOH} = \frac{y}{40} \times 1000 \]

5. **Write the First Equation:**
- From the first titration (20 mL):
\[ \frac{x}{106} + \frac{y}{40} = 10 \]

6. **Calculate the Milliequivalents of NaHCO₃:**
- The milliequivalents of NaHCO₃ reacting with the additional HCl (8 mL):
\[ \text{Milliequivalents of NaHCO}_3 = 4 \]
- Since the NaHCO₃ comes from Na₂CO₃, we have:
\[ \frac{x}{106} = 4 \]

7. **Solve for \( x \):**
- Rearranging gives:
\[ x = 4 \times 106 = 424 \, \text{mg} = 0.424 \, \text{g} \]

8. **Substitute \( x \) back into the First Equation:**
- Substitute \( x \) into the first equation:
\[ \frac{0.424}{106} + \frac{y}{40} = 10 \]
- Calculate \( \frac{0.424}{106} \):
\[ \frac{0.424}{106} = 0.004 \]
- Thus:
\[ 0.004 + \frac{y}{40} = 10 \]
- Rearranging gives:
\[ \frac{y}{40} = 10 - 0.004 = 9.996 \]
- Therefore:
\[ y = 9.996 \times 40 = 399.84 \, \text{mg} = 0.39984 \, \text{g} \]

9. **Calculate the Percentage of Na₂CO₃:**
- The percentage of Na₂CO₃ in the mixture:
\[ \text{Percentage of Na}_2\text{CO}_3 = \left( \frac{x}{0.848} \right) \times 100 = \left( \frac{0.424}{0.848} \right) \times 100 = 50\% \]

### Final Answer:
The percentage of Na₂CO₃ in the mixture is **50%**.