Chemical Engineering - Stoichiometry - Discussion
Discussion Forum : Stoichiometry - Section 2 (Q.No. 39)
39.
The number of atoms of oxygen present in 11.2 litres of ozone (O3) at N.T.P. are
Discussion:
4 comments Page 1 of 1.
Balaram said:
6 years ago
It should be 9.03 * 10^23.
SOHAIL Shahani said:
4 years ago
Can you pls elaborate how to solve this?
Shirish kumar said:
4 years ago
No of mole = volume given/22.4=11.2/22.4 = 0.5,
Avogadro's =6.02*10^23,
6.02*10^23*0.5=3.01*10^23.
FOR
O3 =3*3.01*10^23=9.03*10^23 Ans.
Avogadro's =6.02*10^23,
6.02*10^23*0.5=3.01*10^23.
FOR
O3 =3*3.01*10^23=9.03*10^23 Ans.
(3)
Mehtab Ali said:
3 years ago
1 mole of O3 = 22.4 L = 6.02x10^23 atoms..
0.5 mole of O3 = 11.2 = 3.01x10^23 atoms.
0.5 mole of O3 = 11.2 = 3.01x10^23 atoms.
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