A neon - dioxygen mixture contains 70.6 g dioxygen and 167.5 g neon. If pressure of the mixture of gases in the cylinder is 25 bar. What isd the partial pressure of dioxygen and neon in the mixture ?

Explanation

Calculation of moles of Dioxygen gas,
Where, molecular mass of `O_(2)=32g mol^(-1)`
mole `=("Mass")/("Molecular Mass")=(70.6 g)/(32g mol^(-1))`
`therefore n(O_(2))=2.206 ~~2.21` mol
Caslculation of mole number of Neon gas :
Mole = n (Ne) `= ("Weight")/("Molar Mass")`
`= (167.5 g)/(20 g mol^(-1))=8.375` mol
Total moles of mixture `= n(O_(2))+n (Ne)`
`= (2.21+8.375)` mol
= 10.585 mol
Mole fraction of `O_(2)chi (O_(2))=("Moles of "O_(2))/("Total Moles")`
`= (2.21 mol)/(10.585 mol)=0.2087`
Partial pressure of `O_(2)p(O_(2))`
`= chi (O_(2))xx` total Pressure (P)
`=(2.21mol)/(10.585 mol)xx25` bar
`= 5.2196` bar `~~ 5.22` bar
Partial Pressure of `N_(2)` gas `(P_(N_(2)))`
`= chi_(Ne)xx` total Pressure,
`= 25xx((8.375)/(10.585))`
= 19.7803 bar
OR
`rho_(Ne)=P_("total")-rho_(O_(2))`
`= (25.5.2196)` bar = 19.7804 bar