180 mL of pure phosphine is decomposed to produce vapours of phosphorus and `H_2`. The change in volume during reaction is :

Options

A. 60mL
B. 40mL
C. 90mL
D. 110mL

Explanation

To solve the problem, we need to analyze the decomposition of phosphine (PH₃) and calculate the change in volume during the reaction.

### Step-by-Step Solution:

1. **Write the Decomposition Reaction:**
The decomposition of phosphine can be represented as:
\[ 4 \text{PH}_3 \rightarrow \text{P} + 6 \text{H}_2 \]
This means that 4 moles of phosphine decompose to produce 1 mole of phosphorus and 6 moles of hydrogen.

2. **Determine the Volume of Gases Involved:**
We know that the volume of a gas is directly proportional to the number of moles (at constant temperature and pressure). Therefore, we can assume:
- 4 mL of PH₃ produces 1 mL of P (solid, negligible volume) and 6 mL of H₂.

3. **Calculate the Volume of Hydrogen Produced from 180 mL of PH₃:**
Using the stoichiometric ratios from the balanced equation:
\[ \text{Volume of H}_2 = \left(\frac{6 \text{ mL of H}_2}{4 \text{ mL of PH}_3}\right) \times 180 \text{ mL of PH}_3 \]
\[ \text{Volume of H}_2 = \frac{6}{4} \times 180 = 270 \text{ mL} \]

4. **Calculate the Change in Volume:**
The change in volume during the reaction can be calculated as:
\[ \text{Change in Volume} = \text{Volume of H}_2 - \text{Volume of PH}_3 \]
\[ \text{Change in Volume} = 270 \text{ mL} - 180 \text{ mL} = 90 \text{ mL} \]

5. **Conclusion:**
The change in volume during the decomposition of 180 mL of pure phosphine is **90 mL**.

### Final Answer:
The change in volume during the reaction is **90 mL**.

---