The mole fraction of dioxygen in a neon-dioxygen mixture is 0.18. If the total pressure of the mixture is 25 bar, the partial pressure of neon in the mixture would be

Question

The mole fraction of dioxygen in a neon-dioxygen mixture is 0.18. If the total pressure of the mixture is 25 bar, the partial pressure of neon in the mixture would be (A) 25.18 bar (B) 25.82 bar (C) 4.5 bar (D) 20.5 bar

Tardigrade Admin · Accepted Answer

Given Ptotal = 25 bar, xO2 = 0.18 ∴ : :: xNe = 1 - 0.18 = 0.82 pNe = xNe × Ptotal = 0.82 × 25 = 20.5 bar

Q. The mole fraction of dioxygen in a neon-dioxygen mixture is $0.18$ . If the total pressure of the mixture is $25$ bar, the partial pressure of neon in the mixture would be

$25.18 ba r$

$25.82 ba r$

$4.5 ba r$

$20.5 ba r$

Given $P_{t o t a l}$ = 25 bar,
$x_{O_{2}} = 0.18$
$∴ x_{N e} = 1 - 0.18 = 0.82$
$p_{N e} = x_{N e} \times P_{t o t a l} = 0.82 \times 25 = 20.5 ba r$

Q. The mole fraction of dioxygen in a neon-dioxygen mixture is 0.18. If the total pressure of the mixture is 25 bar, the partial pressure of neon in the mixture would be

25.18bar

25.82bar

4.5bar

20.5bar