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 ?

A neon - dioxygen mixture contains 70.6 g dioxygen and 167.5 g neon. I

1.	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 ?
Answer» Solution :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