The volume occupied by 3.2 g of oxygen gas at S.T.P is:

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.