An orange coloured solution acidified with `H_2SO_4` and treated with a substance `'X'` gives a blue coloured solution of `CrO_5`. The substance `'X'` is

Options

A. `H_(2)O`
B. Dil. HCl
C. `H_(2)O_(2)`
D. Conc. HCl

Answer

To solve the question, we need to identify the substance 'X' that, when added to an orange-colored solution acidified with H₂SO₄, produces a blue-colored solution of CrO₅.

### Step-by-Step Solution:

1. **Identify the Orange-Colored Solution**:
- The orange-colored solution mentioned in the question is likely potassium dichromate (K₂Cr₂O₇). This compound is known for its distinctive orange color.

2. **Acidification with H₂SO₄**:
- When potassium dichromate is acidified with sulfuric acid (H₂SO₄), it forms a solution of dichromic acid (H₂Cr₂O₇). The potassium ions (K⁺) are replaced by hydrogen ions (H⁺) from the sulfuric acid.

\[ K_2Cr_2O_7 + H_2SO_4 \rightarrow H_2Cr_2O_7 + K_2SO_4 \]

3. **Reaction with Substance 'X'**:
- Now, we need to determine what substance 'X' can react with the dichromic acid (H₂Cr₂O₇) to produce CrO₅, which is a blue-colored compound.

4. **Identifying Substance 'X'**:
- The only substance among the options provided that can react with H₂Cr₂O₇ to produce CrO₅ is hydrogen peroxide (H₂O₂). The reaction can be represented as follows:

\[ H_2Cr_2O_7 + H_2O_2 \rightarrow CrO_5 + H_2O \]

5. **Conclusion**:
- Therefore, the substance 'X' that gives a blue-colored solution of CrO₅ when treated with the acidified orange solution is hydrogen peroxide (H₂O₂).

### Final Answer:
The substance 'X' is **H₂O₂ (hydrogen peroxide)**.

---