Excess `F_(2)(g)` reacts at `150^(@)C` and 1.0 atm pressure with `Br_(2)(g)` to give a compound `BrF_(n)`. If 423 mL of `Br_(2)(g)` at the same temperature and pressure produced 4.2 g of `BrF_(n)`, what is n? [Atomic mass Br =80, F =19 ]

Explanation

To solve the problem, we will follow these steps:

### Step 1: Convert the volume of Br₂ gas to liters
Given that the volume of Br₂ is 423 mL, we convert it to liters:
\[ \text{Volume of Br}_2 = \frac{423 \, \text{mL}}{1000} = 0.423 \, \text{L} \]

### Step 2: Convert the temperature to Kelvin
The temperature is given as 150°C. To convert to Kelvin:
\[ T(K) = 150 + 273 = 423 \, K \]

### Step 3: Use the Ideal Gas Law to find moles of Br₂
Using the Ideal Gas Law \( PV = nRT \):
- \( P = 1 \, \text{atm} \)
- \( V = 0.423 \, \text{L} \)
- \( R = 0.0821 \, \text{L atm/(K mol)} \)
- \( T = 423 \, K \)

Rearranging for \( n \):
\[ n = \frac{PV}{RT} = \frac{1 \times 0.423}{0.0821 \times 423} \]
Calculating:
\[ n = \frac{0.423}{34.7063} \approx 0.0122 \, \text{moles of Br}_2 \]

### Step 4: Determine moles of BrFn produced
From the reaction stoichiometry, 1 mole of Br₂ produces 2 moles of BrFn. Therefore:
\[ \text{Moles of BrFn} = 2 \times 0.0122 = 0.0244 \, \text{moles} \]

### Step 5: Calculate the molar mass of BrFn
Given that the mass of BrFn produced is 4.2 g, we can find the molar mass:
\[ \text{Molar mass of BrFn} = \frac{\text{mass}}{\text{moles}} = \frac{4.2 \, \text{g}}{0.0244 \, \text{moles}} \approx 172.48 \, \text{g/mol} \]

### Step 6: Set up the equation for the molar mass of BrFn
The molar mass of BrFn can be expressed as:
\[ \text{Molar mass of BrFn} = 80 + 19n \]
Setting this equal to the calculated molar mass:
\[ 80 + 19n = 172.48 \]

### Step 7: Solve for n
Rearranging the equation:
\[ 19n = 172.48 - 80 \]
\[ 19n = 92.48 \]
\[ n = \frac{92.48}{19} \approx 4.87 \]

Rounding to the nearest whole number, we find:
\[ n \approx 5 \]

### Final Answer
Thus, the value of \( n \) is **5**.

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