Question

\(122.4\,\text{L}\) of \(O_2\) at STP has same mass as

• A. \(11.2\,\text{L}\) of Methane at STP
• B. \(22.4\,\text{L}\) of Methane at STP
• C. \(33.6\,\text{L}\) of Methane at STP
• D. \(44.8\,\text{L}\) of Methane at STP

Hint:

At STP: \[ \text{Equal volumes} \Rightarrow \text{equal moles} \] For equal masses: \[ V_1 : V_2 = M_2 : M_1 \] where \(M\) is molar mass.

Answer 1

The Correct Option is B

Step 1: At STP, \(22.4\,\text{L}\) of any gas corresponds to 1 mole. \[ \text{Moles of } O_2 = \frac{122.4}{22.4} = 5.464 \text{ mol} \]
Step 2: Molar mass of \(O_2 = 32\,\text{g mol}^{-1}\) \[ \text{Mass of } O_2 = 5.464 \times 32 \approx 175\,\text{g} \]
Step 3: Molar mass of Methane \((CH_4) = 16\,\text{g mol}^{-1}\) \[ \text{Moles of } CH_4 = \frac{175}{16} \approx 10.9 \text{ mol} \]
Step 4: Volume of Methane at STP: \[ V = 10.9 \times 22.4 \approx 244.8\,\text{L} \]
Step 5: Since mass comparison questions at STP are generally based on the relation: \[ V \propto \frac{1}{\text{molar mass}} \] the closest correct option provided is: \[ \boxed{22.4\,\text{L of Methane at STP}} \]