A mixture of `Na_(2)CO_(3) and Na_(2)SO_(3)` is treated with dilute `H_(2)SO_(4)` I a setus such that the gaseous mixture emerging can pass first through a solution of `BaCl_(2)` and then gases mixture passed through acidified `K_(2)Cr_(2)O_(7)`. Which of the following will you observe?

Options

A. The `BaCl_(2)` solution remains unaffected and the acidified dichromate soution turns green
B. The `BaCl_(2)` solution gives a white precipitate and the acidfied dichromate solution remains unaffected
C. The `BaCl_(2)` solution givbes a white precipitate and the acidified dichromate solutin turns green
D. Both the solutions remains unaffected

Answer

To solve the problem, we need to analyze the reactions that occur when a mixture of sodium carbonate (Na₂CO₃) and sodium sulfite (Na₂SO₃) is treated with dilute sulfuric acid (H₂SO₄) and how the resulting gases interact with barium chloride (BaCl₂) and acidified potassium dichromate (K₂Cr₂O₇).

### Step-by-Step Solution:

1. **Identify the Components**:
The mixture contains sodium carbonate (Na₂CO₃) and sodium sulfite (Na₂SO₃).

2. **Reaction with Dilute H₂SO₄**:
When the mixture is treated with dilute sulfuric acid:
- Sodium carbonate reacts to form carbon dioxide (CO₂) and sodium sulfate (Na₂SO₄):
\[ \text{Na}_2\text{CO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{CO}_2 + \text{H}_2\text{O} \]
- Sodium sulfite reacts to form sulfur dioxide (SO₂) and sodium sulfate:
\[ \text{Na}_2\text{SO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{SO}_2 + \text{H}_2\text{O} \]
Therefore, the gases produced are carbon dioxide (CO₂) and sulfur dioxide (SO₂).

3. **Passing Through BaCl₂ Solution**:
The gaseous mixture (CO₂ and SO₂) is passed through a solution of barium chloride (BaCl₂):
- Carbon dioxide reacts with barium chloride to form barium carbonate (BaCO₃), which is a white precipitate:
\[ \text{CO}_2 + \text{BaCl}_2 \rightarrow \text{BaCO}_3 \downarrow + \text{HCl} \]
- Sulfur dioxide reacts with barium chloride to form barium sulfite (BaSO₃), which is also a white precipitate:
\[ \text{SO}_2 + \text{BaCl}_2 \rightarrow \text{BaSO}_3 \downarrow + \text{HCl} \]
Both reactions lead to the formation of white precipitates.

4. **Passing Through Acidified K₂Cr₂O₇**:
After passing through BaCl₂, the gaseous mixture is then passed through acidified potassium dichromate (K₂Cr₂O₇):
- In the presence of sulfur dioxide, the dichromate ion (Cr₂O₇²⁻) is reduced to chromium(III) ion (Cr³⁺), which gives a green color:
\[ \text{SO}_2 + \text{K}_2\text{Cr}_2\text{O}_7 + \text{H}_2\text{SO}_4 \rightarrow \text{Cr}^{3+} + \text{other products} \]
- However, since the BaCl₂ solution has already reacted with the gases, the K₂Cr₂O₇ solution remains unaffected by CO₂.

5. **Final Observations**:
- The BaCl₂ solution will show a white precipitate due to the formation of BaCO₃ and BaSO₃.
- The acidified K₂Cr₂O₇ solution will turn green due to the reduction of Cr₂O₇²⁻ by SO₂.

### Conclusion:
The correct observations are:
- BaCl₂ solution gives a white precipitate.
- Acidified K₂Cr₂O₇ solution turns green.