Question

Calculate the actual volume occupied by 1 mol of H2O​

Answer 1

Answer:

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.

PLEASE FOLLOW ME .....

Answer 2

Answer:

Calculate the actual volume of one molecule of water.

ans = 2.99 \times {10}^{ - 23} molans=2.99×10

−23

mol

━━━━━━━━━━━━━━━

Explanation:-

density \: of \: water = 1g {ml}^{ - 1}densityofwater=1gml

−1

1g \: of \: water \: = 1ml1gofwater=1ml

1mole \: of \: water \: = 18g \: of \: h2o1moleofwater=18gofh2o

= 18 \: ml \: of \: water=18mlofwater

Therefore,

6.022 \times {10}^{23} molecules \: of \: water \: occupy \: 18ml6.022×10

23

moleculesofwateroccupy18ml

1molecule \: of \: water \: occupy = \frac{18}{6.022 \times {10}^{23} }1moleculeofwateroccupy=

6.022×10

23

18

= 2.99 \times {10}^{23} mol=2.99×10

23

mol