8.2 L of an ideal gas weights 9.0 g at 300 K and 1 atm pressure. The molecular mass of the gas is

Options

A. 54
B. 27
C. 13.5
D. 81

Explanation

To find the molecular mass of the ideal gas, we can use the ideal gas equation and the relationship between mass, moles, and molecular mass. Here’s a step-by-step solution:

### Step 1: Write down the ideal gas equation.
The ideal gas equation is given by:
\[ PV = nRT \]
where:
- \( P \) = pressure (in atm)
- \( V \) = volume (in liters)
- \( n \) = number of moles
- \( R \) = universal gas constant (0.0821 atm·L/(mol·K))
- \( T \) = temperature (in Kelvin)

### Step 2: Substitute the known values into the ideal gas equation.
Given:
- \( P = 1 \, \text{atm} \)
- \( V = 8.2 \, \text{L} \)
- \( T = 300 \, \text{K} \)

Substituting these values into the equation:
\[ 1 \, \text{atm} \times 8.2 \, \text{L} = n \times 0.0821 \, \text{atm·L/(mol·K)} \times 300 \, \text{K} \]

### Step 3: Solve for the number of moles \( n \).
Rearranging the equation to solve for \( n \):
\[ n = \frac{PV}{RT} \]
\[ n = \frac{1 \, \text{atm} \times 8.2 \, \text{L}}{0.0821 \, \text{atm·L/(mol·K)} \times 300 \, \text{K}} \]

Calculating \( n \):
\[ n = \frac{8.2}{24.63} \approx 0.333 \, \text{moles} \]

### Step 4: Use the relationship between mass, moles, and molecular mass.
The relationship is given by:
\[ n = \frac{\text{mass}}{\text{molecular mass}} \]
Rearranging this gives:
\[ \text{molecular mass} = \frac{\text{mass}}{n} \]

### Step 5: Substitute the known values to find the molecular mass.
Given:
- Mass of the gas = 9.0 g
- Number of moles \( n \) = 0.333 moles

Substituting these values:
\[ \text{molecular mass} = \frac{9.0 \, \text{g}}{0.333 \, \text{moles}} \]

Calculating the molecular mass:
\[ \text{molecular mass} \approx 27 \, \text{g/mol} \]

### Final Answer:
The molecular mass of the gas is approximately **27 g/mol**.

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