Pure phosphine originally present at 2.5 atm and 300 K decomposes slowly according to the equation:
`4PH_3(g) iff P_4(g) + 6H_2(g)`
What is the vapour density of phosphine if it dissociates to the extent of 40%?

Explanation

To solve the problem of finding the vapor density of phosphine (PH₃) after it dissociates to the extent of 40%, we will follow these steps:

### Step 1: Understand the Reaction
The decomposition of phosphine is given by the equation:
\[ 4 \text{PH}_3(g) \iff \text{P}_4(g) + 6 \text{H}_2(g) \]

### Step 2: Initial Conditions
We start with pure phosphine at a pressure of 2.5 atm and a temperature of 300 K. The initial number of moles of PH₃ can be represented as \( n_0 \).
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