calculate the volume occupied by 28 u nitrogen gas.
Answers
Answered by
841
molecular mass of nitrogen gas = 28u
28u nitrogen gas contains 1 molecule of nitrogen.
1 mole of nitrogen gas at STP occupies = 22.4 l
or 6.022*10^23 molecules of nirogen occupies = 22.4l
1 molecule of nirtogen occupies = 22.4l/6.022*10^23 = 3.72*10^(-23) litre
So 28u of nitrogen will occupy 3.72*10^(-23) litre
28u nitrogen gas contains 1 molecule of nitrogen.
1 mole of nitrogen gas at STP occupies = 22.4 l
or 6.022*10^23 molecules of nirogen occupies = 22.4l
1 molecule of nirtogen occupies = 22.4l/6.022*10^23 = 3.72*10^(-23) litre
So 28u of nitrogen will occupy 3.72*10^(-23) litre
Answered by
147
Answer:
28u =1 mole of N2
then volume occupied is 22.4L
so, 22.4L/6.022*10^23=3.72*10^23.
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