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At 273 K, the density of a gaseous oxide...

At 273 K, the density of a gaseous oxide is 2 bar is same as that of nitrogen at 5 bar. The molecular mass of the oxide is

A

`50 g " mol"^(-1)`

B

`90 g " mol"^(-1)`

C

`70 g " mol"^(-1)`

D

`120g " mol"^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
C
`d=(PM)/(RT)`
For nitrogen gas at 5 bar pressure and 273 K temperature ,
`d(N_2)=( 5 " bar " xx 28 g "mol"^(-1))/(R xx 273 K)`
Now , `d(N_2) = ` d(oxide)
`(5"bar " xx 28 g " mol"^(-1))/( R xx 273 K) = (2 "bar" xx M g "mol"^(-1))/( R xx 273K)`
`:. M =(5 xx 28)/(2) = 70 g "mol"^(-1)`
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