`34.05 mL` of phosphorus vapours weighs `0.0625 g` at `546^(@)C` and `0.1` bar pressure. What is the molar mass of phosphorus ?

Explanation

To find the molar mass of phosphorus using the given data, we will follow these steps:

### Step 1: Identify the given values
- Volume (V) = 34.05 mL
- Mass (m) = 0.0625 g
- Temperature (T) = 546 °C
- Pressure (P) = 0.1 bar

### Step 2: Convert the volume from mL to L
To convert mL to L, we use the conversion factor \(1 \text{ L} = 1000 \text{ mL}\):
\[ V = 34.05 \text{ mL} = \frac{34.05}{1000} = 0.03405 \text{ L} \]

### Step 3: Convert the temperature from Celsius to Kelvin
To convert Celsius to Kelvin, we add 273.15:
\[ T = 546 + 273.15 = 819.15 \text{ K} \]

### Step 4: Use the Ideal Gas Law to find the molar mass
The Ideal Gas Law is given by the equation:
\[ PV = \frac{m}{M}RT \]
Where:
- \(P\) = Pressure in bar
- \(V\) = Volume in L
- \(m\) = Mass in g
- \(M\) = Molar mass in g/mol
- \(R\) = Ideal gas constant = 0.0821 L·bar/(K·mol)
- \(T\) = Temperature in K

Rearranging the formula to solve for molar mass \(M\):
\[ M = \frac{mRT}{PV} \]

### Step 5: Substitute the values into the equation
Substituting the known values into the rearranged equation:
\[ M = \frac{0.0625 \text{ g} \times 0.0821 \text{ L·bar/(K·mol)} \times 819.15 \text{ K}}{0.1 \text{ bar} \times 0.03405 \text{ L}} \]

### Step 6: Calculate the molar mass
Now, we perform the calculations step-by-step:
1. Calculate the numerator:
\[ 0.0625 \times 0.0821 \times 819.15 = 4.066 \]
2. Calculate the denominator:
\[ 0.1 \times 0.03405 = 0.003405 \]
3. Now divide the numerator by the denominator:
\[ M = \frac{4.066}{0.003405} \approx 119.5 \text{ g/mol} \]

### Final Result
The molar mass of phosphorus is approximately **119.5 g/mol**.

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