If `C_(a_(3))(PO_(4))_(2) and H_(3)PO_(3)` contain same number of 'P' atom then the ratio of oxygen atoms in these compounds respectively is

Options

A. `8//3`
B. `2//3`
C. 3
D. `4//3`

Answer

To solve the problem, we need to determine the ratio of oxygen atoms in the compounds \( \text{Ca}_3(\text{PO}_4)_2 \) and \( \text{H}_3\text{PO}_3 \), given that they contain the same number of phosphorus (P) atoms.

### Step-by-Step Solution:

1. **Identify the number of P atoms in each compound**:
- In \( \text{Ca}_3(\text{PO}_4)_2 \):
- The formula indicates that there is 1 P atom in each \( \text{PO}_4 \) unit.
- Since there are 2 \( \text{PO}_4 \) units, the total number of P atoms is:
\[ 2 \text{ P atoms} \]
- In \( \text{H}_3\text{PO}_3 \):
- The formula indicates that there is 1 P atom in \( \text{H}_3\text{PO}_3 \).
- Therefore, the total number of P atoms is:
\[ 1 \text{ P atom} \]

2. **Set the number of P atoms equal**:
- According to the problem, both compounds contain the same number of P atoms:
\[ 2x = y \]
- Here, \( x \) is the number of moles of \( \text{Ca}_3(\text{PO}_4)_2 \) and \( y \) is the number of moles of \( \text{H}_3\text{PO}_3 \).

3. **Determine the number of O atoms in each compound**:
- For \( \text{Ca}_3(\text{PO}_4)_2 \):
- Each \( \text{PO}_4 \) has 4 O atoms, and there are 2 \( \text{PO}_4 \) units:
\[ \text{O atoms} = 2 \times 4 = 8 \text{ O atoms} \]
- For \( \text{H}_3\text{PO}_3 \):
- Each \( \text{H}_3\text{PO}_3 \) has 3 O atoms:
\[ \text{O atoms} = 3 \text{ O atoms} \]

4. **Express the total number of O atoms in terms of moles**:
- For \( \text{Ca}_3(\text{PO}_4)_2 \):
- Total O atoms = \( 8x \)
- For \( \text{H}_3\text{PO}_3 \):
- Total O atoms = \( 3y \)

5. **Substitute \( y \) in terms of \( x \)**:
- From step 2, we have \( y = 2x \).
- Substitute \( y \) into the O atoms expression for \( \text{H}_3\text{PO}_3 \):
\[ \text{O atoms in } \text{H}_3\text{PO}_3 = 3(2x) = 6x \]

6. **Calculate the ratio of O atoms**:
- The ratio of O atoms in \( \text{Ca}_3(\text{PO}_4)_2 \) to \( \text{H}_3\text{PO}_3 \) is:
\[ \text{Ratio} = \frac{8x}{6x} = \frac{8}{6} = \frac{4}{3} \]

### Final Answer:
The ratio of oxygen atoms in \( \text{Ca}_3(\text{PO}_4)_2 \) to \( \text{H}_3\text{PO}_3 \) is \( \frac{4}{3} \).