General Knowledge - Chemistry - Discussion
Discussion Forum : Chemistry - Section 1 (Q.No. 56)
56.
The number of g-molecule of oxygen in 6.02 x 1024CO molecules is
Discussion:
25 comments Page 3 of 3.
Tushar Verma said:
8 years ago
Option C is the absolutely perfect answer. I agree.
Rahul said:
8 years ago
Can anyone explain me this question properly? Please!
Anshika said:
7 years ago
C is the correct answer.
CO has 16g O. i.e. half the amount of O2 (32g).
Now. Moles will be 6.022 * 10^24 * 1/2 ÷ 6.022 * 10^23.
i.e. equal to 5 moles (gm molecule).
CO has 16g O. i.e. half the amount of O2 (32g).
Now. Moles will be 6.022 * 10^24 * 1/2 ÷ 6.022 * 10^23.
i.e. equal to 5 moles (gm molecule).
Usman said:
7 years ago
When we have one gram then why we need mole?
Irine said:
6 years ago
2C + O2 = 2CO.
2mol of CO = 1 mol of O2,
2 * 6.02 * 10^23 molecules of CO= 1g molecule of O2( bcs 1 mol = 6.02 * 10^23 molecules)
6.02 * 10^24 molecules of CO= ?
= 6.02 * 10^24/ 2,
= 6.02 *10^23.
= 10/2= 5 g molecules of O2.
2mol of CO = 1 mol of O2,
2 * 6.02 * 10^23 molecules of CO= 1g molecule of O2( bcs 1 mol = 6.02 * 10^23 molecules)
6.02 * 10^24 molecules of CO= ?
= 6.02 * 10^24/ 2,
= 6.02 *10^23.
= 10/2= 5 g molecules of O2.
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