Question

at NTP
If all the O-atoms from 4.4 g CO 6.022 * 102 molecules of N,0.0.2 moles of CO and 1.12 L of so gas at NTP
are removed and combined to form o gas, then the resulting gas occupies a volume of
(A) 22.4L
(B) 44.8L
(C) 33 6L
(D) 112L​

Answer 1

Answer:

The volume occupied by the 1mole of N

2

=22.4l

So, the 1mole=6.022×10

23

molecules of N

2

=

6.022×10

23

22.4

So, 6.022×10

22

molecules of N

2

would occupy volume

=

6.022×10

23

l

22.4

×6.022×10

22

=2.24l

Thus ,the right option is B