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

Correct Answer: B. 2.24dm3

At Standard Temperature and Pressure (STP), the conditions are defined as:
• Temperature (T) = 273.15 Kelvin (0 degrees Celsius)
• Pressure (P) = 1 atmosphere (atm) or 101.325 kilopascals (kPa)
To find the volume occupied by a gas at STP, you can use the ideal gas law:
PV = nRT
Where:
• P is the pressure (in atmospheres)
• V is the volume (in liters)
• n is the number of moles of the gas
• R is the ideal gas constant (approximately 0.0821 L.atm/(mol.K))
• T is the temperature (in Kelvin)
• First, we need to find the number of moles (n) of oxygen gas (O2) using its molar mass:
• Molar mass of O2 = 32 g/mol
• Given mass of O2 = 3.2 g
• n = (Given mass) / (Molar mass) n = 3.2 g / 32 g/mol n = 0.1 mol
• Now, we can use the ideal gas law to find the volume (V) at STP:
• P = 1 atm T = 273.15 K
• n = 0.1 mol R = 0.0821 L.atm/(mol.K)
• PV = nRT
• V = (nRT) / P V = (0.1 mol * 0.0821 L.atm/(mol.K) * 273.15 K) / 1 atm V ≈ 2.24 L
• So, the volume occupied by 3.2 g of oxygen gas at STP is approximately 2.24 liters.
• Therefore, the correct answer is:
• B. 2.24 dm3 (which is equivalent to 2.24 liters)