what is the answer of above question?
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Answer:
2.<1% of 22.4 L
Explanation:
At STP, volume occupied by one mole of gas is 22.4 L.
However, 1 mole H2O = 18 gm
Now, assuming density of water equal to 1,
18 gm H2O=18 ml H20
% of actual volume occupied= Actual volume x 100/ Volume of water vapour at STP
= 18x100/22400 (22.4 L= 22400 ml)
= 0.08%
so < 1% of 22.4 L is the Answer
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