A mixture of He and `O_(2)` has density `1.3 gm//`litre at NTP. Them mole fraction of He is
A. `0.1`
B. `0.9`
C. `0.4`
D. `0.6`

5 views

1 Answers

Correct Answer - A
`d = 1.3 gm//"litre" = (PM)/(RT)`
`1.3 = (1 xx M)/(0.08 xx 273)`
`M = 1.3 xx 22.4`
`M = 29.12 gm//mol`
`M_(avg) = M_(1)X_(1) xx M_(2) X_(2)`
`29.12 = 4a + 32 - 32a`
`29.12 = 32 - 28a`
`28a = 32 - 29.12`
`a = (2.88)/(28)`
`a = 0.1` Hence, `X_(O_(2)) = 0.9`
`X_(He) = 0.1`

5 views

Related Questions