Assuming the density of water to be 1 g/ml, the volume occupied by one molecule of water is
Answers
Hey mate here's your answer
The number of moles in 1 litre = 1000 / 18.02 = 55.5 moles. 1 mole of water = 6.022 x 10^23 molecules of water, So 55.5 moles of water = 55.5 x 6.022 x 10^23 = 3.34 x 10^25 molecules of water.
The gram molecular mass of water is 18.02. Thus volume occupied by 1 H2O molecule is 2.99 ×10-23 ml.
Answer:
Explanation:
The density of water is approximately 1g/ml.
Thus, 1g of water occupy volume = 1 ml
The molar mass of water = 18 g/mol
Number of moles of water in 1g =
18
1
moles = 0.055 moles H
2
O
Now 18 g water contain = 6.022 × 10
23
moleculesH
2
O
Thus 1 g water will have =
18
6.022×10
23
molecules H
2
O = 0.334 × 10
23
molecules H
2
O
This shows that
0.334 ×10
23
molecules H
2
O will occupy volume = 1 ml
Therefore volume occupied by 1 H
2
O molecule =
0.334×10
23
1
=2.99 × 10
−23
ml
Thus volume occupied by 1 H
2
O mlecule is 2.99 × 10
−23
ml