A1.00 l sample of a gas at 25.0◦c and 1.00 atm contains 50.0 % helium and 50.0 % neon by mass. what is the partial pressure of the neon. express your answer in atmospheres.

V = Volume of sample of gas = 1.00 L = 0.001 m³

T = temperature of the gas = 25.0 ⁰C = 25 + 273 = 298 K

P = pressure = 1.00 atm = 101325 Pa

n = number of moles of gas

using ideal gas equation

PV = n RT

101325 (0.001) = n (8.314) (298)

n = 0.041

n₁ = number of moles of helium

n₂ = number of moles of neon

m₁ = mass of helium = n₁ (4) = 4 n₁

m₂ = mass of neon = n₂ (20.2) = 20.2 n₂

given that :

m₁ = m₂

4 n₁ = 20.2 n₂

n₁ = 5.05 n₂

also

n₁ +  n₂ = n

5.05 n₂ +  n₂ = 0.041

n₂ = 0.0068

mole fraction of neon is given as

mole fraction = n₂ /n = 0.0068/0.041 = 0.166

P₂ = partial pressure of neon = (mole fraction) P

P₂ = (0.166) (1)

P₂ = 0.166 atm

answer: b.

explanation:

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