At `27^@C` ina 10L flask 4.0g of an ideal gaseous mixture conatining He (molar mass 4.0 g `mol^-1`) and Ne (molar mass `20g mol^-1`) has a pressure of 1.23 atm. What is the mass% of neon? (`R = 0.082 L atm K^-1 mol^-1)`

Options

A. `25.2`
B. `62.5`
C. `84.2`
D. `74.2`

Answer

B

Explanation

`T = 27 + 273 = 300 K`
v = 10 L
`W = W_(1) W_(2) = 4g `
`M_((He))= 4`
`M_(("Ne")) = 20`
p = 1.23 atm
`r = 0.0 82 L atm K^(-1) mol^(-1)`
(i) `:. pV = (n_(1) +n_(2)) RT`
`= [(W_(1))/(4) +(W)/(20)]` RT
`:. [(W_(1))/(4) +(W_(2))/(20)] = (pV)/(RT) = (1.23 xx 10)/( 0.083 xx 300) = 0.5`
`25 W_(1) + 5 W_(2)= 50`
Or, `5W_(1) +W_(2) = 10 " ". . . (i)`
Also (ii) `W_(1) + w_(2) = 4 " " . . . (ii)`
On subtracting Eq. (ii) from Eq. (i)
`4 W_(1) = 6 rArr W_(1) = 1.5`
and `W_(2) = 4 - 1.5 = 2.5`
Hence, Mass % of Ne (neon)
` = (2.5 xx 10)/(4) = 62 . 5%`