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2.8 g of N(2) , 0.40 g of H(2) and 6.4g ...

2.8 g of `N_(2)` , 0.40 g of `H_(2)` and 6.4g of `O_(2)` are placed in a container of 1.0 L capacity at `27^(@)C`. The total pressure in the container is `:`

A

6.12 atm

B

12.3 atm

C

1.23atm

D

24.6 atm

Text Solution

Verified by Experts

The correct Answer is:
B
Total moles `= ( 2.8 ) /( 28) + ( 0.4)/( 2) + ( 6.4)/( 32) = 0.5 `
`p = ( n RT )/( V )`
or `p = ( 0.5 xx 0.082 xx 300)/( 1)`
`= 12.3 atm`
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