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

1 answer:

3 0

Answer is: <span>A. 18.02 g/mol.

At standard temperature and pressure 1 mol of gas occupied 22.4 liters:
V(H</span>₂O) = 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.

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