When `K_(2)Cr_(2)O_(7)` mixed with `H_(2)SO_(4)` and thorooughly shaken with `H_(2)O_(2)` in presence of ether. Then the floated blue coloured complex is 'X'. The percentage of Cr in the complex is

Options

A. 49.4
B. 29.4
C. 39.4
D. 59.4

Explanation

To find the percentage of chromium in the complex CrO5 formed from the reaction of K2Cr2O7, H2SO4, and H2O2 in the presence of ether, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction involves potassium dichromate (K2Cr2O7), sulfuric acid (H2SO4), and hydrogen peroxide (H2O2) in the presence of ether. The product formed is chromium pentoxide (CrO5).

2. **Determine the Molar Mass of CrO5**:
- The molar mass of chromium (Cr) is approximately 52 g/mol.
- The molar mass of oxygen (O) is approximately 16 g/mol.
- In CrO5, there is 1 chromium atom and 5 oxygen atoms.
- Calculate the total molar mass of CrO5:
\[ \text{Molar mass of CrO5} = 1 \times \text{(molar mass of Cr)} + 5 \times \text{(molar mass of O)} \]
\[ = 1 \times 52 + 5 \times 16 = 52 + 80 = 132 \text{ g/mol} \]

3. **Calculate the Percentage of Chromium in CrO5**:
- The formula for calculating the percentage of an element in a compound is:
\[ \text{Percentage of Cr} = \left( \frac{\text{mass of Cr}}{\text{mass of CrO5}} \right) \times 100 \]
- Substitute the values:
\[ \text{Percentage of Cr} = \left( \frac{52}{132} \right) \times 100 \]
- Calculate:
\[ \text{Percentage of Cr} = 0.3939 \times 100 \approx 39.39\% \]

4. **Final Result**:
The percentage of chromium in the complex CrO5 is approximately 39.4%.