To determine the molar mass of the gas, we can use the Ideal Gas Law, which is given by the equation:
\[ PV = nRT \]
Where:
- \( P \) = pressure (in atm)
- \( V \) = volume (in liters)
- \( n \) = number of moles of the gas
- \( R \) = ideal gas constant (0.0821 L·atm/(mol·K))
- \( T \) = temperature (in Kelvin)
We can rearrange the equation to solve for \( n \):
\[ n = \frac{PV}{RT} \]
Given:
- \( P = 0.95 \) atm
- \( V = 4.92 \) L
- \( R = 0.0821 \) L·atm/(mol·K)
- \( T = 280 \) K
Substituting the values into the equation:
\[ n = \frac{(0.95 , \text{atm}) \times (4.92 , \text{L})}{(0.0821 , \text{L·atm/(mol·K)}) \times (280 , \text{K})} \]
Calculating the denominator:
\[ 0.0821 \times 280 = 22.996 \]
Now calculate \( n \):
\[ n = \frac{(0.95) \times (4.92)}{22.996} \approx \frac{4.674}{22.996} \approx 0.203 \]
Now that we have the number of moles \( n \), we can calculate the molar mass \( M \) using the following formula:
\[ M = \frac{m}{n} \]
Where \( m \) is the mass of the gas (5.49 g).
Calculating the molar mass:
\[ M = \frac{5.49 , \text{g}}{0.203 , \text{mol}} \approx 27.0 , \text{g/mol} \]
Thus, the molar mass of the gas is approximately 27 g/mol.
The correct response is:
27 g/mol