The volume occupied by 3.2 g of oxygen gas at S.T.P is:
Correct answer: B. 2.24 dm3
- A. 1.12 dm3
- B. 2.24 dm3
- C. 22.4 dm3
- D. 24 dm3
Explanation
Option B is correct. The volume occupied by 3.2 g of oxygen gas at STP is 2.24 dm3. STP stands for standard temperature and pressure, defined as 0 °C (273.15 K) and 1 atm (101325 Pa). To find the volume of a gas at STP, we use the ideal gas law: PV = nRT. Here, P is the pressure in atm, V is the volume in L, n is the number of moles, R is the ideal gas constant (0.08206 L atm mol-1 K-1), and T is the temperature in K.Given 3.2 g of oxygen, we calculate the moles of oxygen using its molar mass (32 g/mol): n = m / M = 3.2 g / 32 g/mol = 0.1 mol. At STP, we plug the values into the ideal gas law:V = nRT / P = (0.1 mol)(0.08206 L atm mol-1 K-1)(273.15 K) / (1 atm) = 2.24 L. Thus, 2.24 L equals 2.24 dm3.Options A, C, and D are incorrect as they either miscalculate the volume based on the number of moles or present incorrect standard gas volumes.
This question appeared on the KMU MDCAT 2023 paper, which you can sit online with every answer explained.
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About Standard Temperature and Pressure
Standard temperature and pressure provide reference conditions for gas calculations: 0°C or 273 K and, in the usual school convention, 1 atm pressure. Questions use these conditions to relate pressure, volume, temperature and amount of gas, including the molar volume of an ideal gas, about 22.4 dm³.
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