0.30 g of gas was found to occupy a volume of 82.0 mL at `27^(@)C` and 3 atm. Pressure. The molecular mass of the gas is

Options

A. 60
B. 30
C. 90
D. unpredictable

Explanation

To find the molecular mass of the gas, we will use the Ideal Gas Law equation, which is given by:

\[ PV = nRT \]

Where:
- \( P \) = Pressure (in atm)
- \( V \) = Volume (in liters)
- \( n \) = Number of moles of the gas
- \( R \) = Ideal gas constant = 0.0821 L·atm/(K·mol)
- \( T \) = Temperature (in Kelvin)

We know that the number of moles \( n \) can also be expressed as:

\[ n = \frac{\text{mass}}{\text{molar mass}} \]

From this, we can rearrange the Ideal Gas Law to find the molar mass:

\[ \text{Molar mass} = \frac{\text{mass} \cdot R \cdot T}{P \cdot V} \]

### Step-by-step Calculation:

1. **Convert the given values:**
- Mass of gas = 0.30 g
- Volume = 82.0 mL = \( 82.0 \times 10^{-3} \) L = 0.082 L
- Temperature = 27°C = 27 + 273 = 300 K
- Pressure = 3 atm

2. **Substitute the values into the formula:**
\[ \text{Molar mass} = \frac{0.30 \, \text{g} \times 0.0821 \, \text{L·atm/(K·mol)} \times 300 \, \text{K}}{3 \, \text{atm} \times 0.082 \, \text{L}} \]

3. **Calculate the numerator:**
\[ 0.30 \times 0.0821 \times 300 = 7.389 \]

4. **Calculate the denominator:**
\[ 3 \times 0.082 = 0.246 \]

5. **Now, divide the numerator by the denominator:**
\[ \text{Molar mass} = \frac{7.389}{0.246} \approx 30.0 \, \text{g/mol} \]

Thus, the molecular mass of the gas is approximately **30 g/mol**.

### Final Answer:
The molecular mass of the gas is **30 g/mol**.