Question

What is the mass of 22.4 L of H2O at STP?

A. 18.02 g/mol
B. 1.24 g/mol
C. 24 g/mol
D. 100 g/mol

Answer 1

Answer is: A. 18.02 g/mol.

At standard temperature and pressure 1 mol of gas occupied 22.4 liters:
V(H₂O) = 22.4 L; volume of water.
Vm = 22.4 L/mol; molar volume at STP.
n(H₂O) = V ÷ Vm.
n(H₂O) = 22.4 L ÷ 22.4 L/mol.
n(H₂O) = 1 mol; amount of substance (water).
M(H₂O) = Ar(O) + 2Ar(H) · g/mol.
M(H₂O) = 16 + 2 ·1.01 · g/mol.
M(H₂O) = 18.02 g/mol; molar mass of water.