At NTPIf all the O-atoms from 4.4 g CO 6.022 * 102 molecules of N,0.0.

1.	At NTPIf 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 NTPare 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» 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

At NTPIf 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 NTPare 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»

The volume occupied by the 1mole of N

=22.4l

So, the 1mole=6.022×10

molecules of N

6.022×10

22.4

So, 6.022×10

molecules of N

would occupy volume

6.022×10

22.4

×6.022×10

=2.24l

Thus ,the RIGHT option is B