To find the molar mass of the gas, we can use the Ideal Gas Law, which is given by the equation:
\[ PV = nRT \]
Where:
- \( P \) is the pressure in atm
- \( V \) is the volume in liters
- \( n \) is the number of moles of gas
- \( R \) is the ideal gas constant (0.0821 L·atm/(K·mol))
- \( T \) is the temperature in Kelvin
Since we want to find the molar mass, we can express the number of moles \( n \) in terms of the mass of the gas and its molar mass \( M \):
\[ n = \frac{m}{M} \]
Substituting this into the ideal gas equation:
\[ PV = \frac{m}{M}RT \]
Rearranging for \( M \):
\[ M = \frac{mRT}{PV} \]
Now, plug in the given values:
- \( m = 5.49 , \text{g} \)
- \( R = 0.0821 , \text{L·atm/(K·mol)} \)
- \( T = 280 , \text{K} \)
- \( P = 0.95 , \text{atm} \)
- \( V = 4.92 , \text{L} \)
Now we can calculate \( M \):
\[ M = \frac{(5.49 , \text{g}) \times (0.0821 , \text{L·atm/(K·mol)}) \times (280 , \text{K})}{(0.95 , \text{atm}) \times (4.92 , \text{L})} \]
Calculating the numerator:
\[ 5.49 \times 0.0821 \times 280 = 128.99908 , \text{g·L·atm/(K·mol)} \]
Calculating the denominator:
\[ 0.95 \times 4.92 = 4.674 , \text{atm·L} \]
Now divide the numerator by the denominator:
\[ M = \frac{128.99908}{4.674} \approx 27.56 , \text{g/mol} \]
Thus, rounding to two decimal places, the molar mass of the gas is approximately:
\[ M \approx 27 , \text{g/mol} \]
According to the options provided, the closest answer is:
27 g/mol.