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A container of 2.461 litre at 27^(@)C ha...

A container of 2.461 litre at `27^(@)C` has a mixture of 0.3 mole of `N_2`, 0.5 mole of He and 6.2 mole of `O_2`. What is the partial pressure of N?

A

70 atm

B

7 atm

C

23.3 atm

D

3 atm

Text Solution

Verified by Experts

The correct Answer is:
D
Total moles `=n(N_2) + n_(He) + n_(O_2) = 0.3 + 0.5 + 6.2 = 7`
`PV = nRT`
`:. P= (7 xx 0.0821 xx 300)/(2.461) = 70.06 atm ~~ 70`
Partrial pressure =Total pressure `xx` mole fraction
`p_(N_2) = 70 xx (0.3)/(7) = 3 atm`
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