Chemical Engineering - Stoichiometry - Discussion
Discussion Forum : Stoichiometry - Section 6 (Q.No. 23)
23.
The number of water molecules present in a drop of water weighing 0.018 gm is 6.023 x __________
Discussion:
2 comments Page 1 of 1.
Ramsajan said:
2 years ago
Molecules of water 0.018/18 = .001 * 6.023.
=.006023*10^23 or 6.023*10^20,
And number of atoms is 18.069 * 10^20.
Because 1 molecule of H20 contains 2 atoms of H2 and 1 atom of O total of 3 atoms.
=.006023*10^23 or 6.023*10^20,
And number of atoms is 18.069 * 10^20.
Because 1 molecule of H20 contains 2 atoms of H2 and 1 atom of O total of 3 atoms.
Kapil said:
7 years ago
Water molecular weight 18 gm.
18gm= 6.023 * 10`23 atom,
So, .018gm =6.023 * 10 '20 atom.
18gm= 6.023 * 10`23 atom,
So, .018gm =6.023 * 10 '20 atom.
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