What volume of beaker contains exactly 2.23x10^-2 mol of nitrogen gas at STP?

V = 0.5 L

Explanation:

Given dа ta:

Moles of nitrogen = 2.23*10⁻² mol (0.0223 mol)

Temperature = 273 K

Pressure = 1 atm

Volume = ?

Solution:

PV = nRT

V = nRT / P

V = 0.0223 mol * 0.0821 atm. mol⁻¹. L. k⁻¹ * 273 K / 1 atm

V = 0.5 L