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 number of oxygen atoms in the compounds \( Ca_3(PO_4)_2 \) and \( H_3PO_3 \) and then find the ratio of these oxygen atoms, given that both compounds contain the same number of phosphorus (P) atoms.

### Step-by-Step Solution:

1. **Identify the number of phosphorus atoms in each compound**:
- In \( Ca_3(PO_4)_2 \):
- Each formula unit contains 2 phosphorus atoms (since there are 2 phosphate groups).
- In \( H_3PO_3 \):
- Each formula unit contains 1 phosphorus atom.

2. **Set the condition for the number of phosphorus atoms**:
- Let \( x \) be the number of moles of \( Ca_3(PO_4)_2 \).
- The total number of phosphorus atoms in \( Ca_3(PO_4)_2 \) is \( 2x \).
- Let \( y \) be the number of moles of \( H_3PO_3 \).
- The total number of phosphorus atoms in \( H_3PO_3 \) is \( y \).
- According to the problem, we have:
\[ 2x = y \]

3. **Determine the number of oxygen atoms in each compound**:
- In \( Ca_3(PO_4)_2 \):
- Each phosphate group \( (PO_4) \) contains 4 oxygen atoms.
- Since there are 2 phosphate groups, the total number of oxygen atoms is:
\[ 4 \times 2x = 8x \]
- In \( H_3PO_3 \):
- Each molecule contains 3 oxygen atoms.
- Therefore, the total number of oxygen atoms is:
\[ 3y \]

4. **Substitute \( y \) in terms of \( x \)**:
- From the equation \( y = 2x \), substituting this into the expression for oxygen in \( H_3PO_3 \):
\[ 3y = 3(2x) = 6x \]

5. **Calculate the ratio of oxygen atoms**:
- Now we have the total number of oxygen atoms in both compounds:
- Oxygen in \( Ca_3(PO_4)_2 \): \( 8x \)
- Oxygen in \( H_3PO_3 \): \( 6x \)
- The ratio of the number of oxygen atoms in \( Ca_3(PO_4)_2 \) to \( H_3PO_3 \) is:
\[ \text{Ratio} = \frac{8x}{6x} = \frac{8}{6} = \frac{4}{3} \]

### Final Answer:
The ratio of oxygen atoms in \( Ca_3(PO_4)_2 \) to \( H_3PO_3 \) is \( \frac{4}{3} \).