To an acidified dichromate solution, a pinch of `Na_(2)O_(2)` is added and shaken.
`Na_(2)Cr_(2)O_(7) +4Na_(2)O_(2)+5H_(2)SO_(4) to 2(A)+5(B)+5H_(2)O`.
Compound (A) is blue coloured
The sum of the total number of atoms in one molecule of (A) and (B) is

Explanation

To solve the given problem, we need to analyze the reaction and identify the compounds (A) and (B) formed, and then calculate the total number of atoms in one molecule of each compound.

### Step-by-Step Solution:

1. **Identify the Reactants and Products**:
The reaction provided is:
\[ \text{Na}_2\text{Cr}_2\text{O}_7 + 4 \text{Na}_2\text{O}_2 + 5 \text{H}_2\text{SO}_4 \rightarrow 2 \text{A} + 5 \text{B} + 5 \text{H}_2\text{O} \]
Here, we know that (A) is a blue-colored compound and (B) is another compound we need to identify.

2. **Identify Compound (A)**:
When sodium peroxide (\(\text{Na}_2\text{O}_2\)) is added to acidified dichromate solution, it leads to the formation of chromium pentoxide (\(\text{CrO}_5\)), which is blue in color. Therefore, we can conclude:
\[ \text{A} = \text{CrO}_5 \]

3. **Identify Compound (B)**:
The other product formed in the reaction is sodium sulfate, which can be represented as:
\[ \text{B} = \text{Na}_2\text{SO}_4 \]

4. **Count the Atoms in Compound (A)**:
For \(\text{CrO}_5\):
- 1 atom of chromium (Cr)
- 5 atoms of oxygen (O)
- Total atoms in (A) = \(1 + 5 = 6\)

5. **Count the Atoms in Compound (B)**:
For \(\text{Na}_2\text{SO}_4\):
- 2 atoms of sodium (Na)
- 1 atom of sulfur (S)
- 4 atoms of oxygen (O)
- Total atoms in (B) = \(2 + 1 + 4 = 7\)

6. **Calculate the Sum of Total Atoms**:
Now, we need to find the sum of the total number of atoms in one molecule of (A) and (B):
\[ \text{Total atoms} = \text{Atoms in (A)} + \text{Atoms in (B)} = 6 + 7 = 13 \]

### Final Answer:
The sum of the total number of atoms in one molecule of (A) and (B) is **13**.

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