Question

A mixture of krypton and oxygen gases at a total pressure of 912 mm Hg contains krypton at a partial pressure of 323 mm Hg. If the gas mixture contains 8.08 grams of krypton, how many grams of oxygen are present?

Mass =
g

Answer 1

Grams O₂ = 5.60 gms

Step-by-step explanation:

Given gas mixture of Kr and O₂

Partial Pressure Kr = P(Kr) = 323mm

Total Pressure (given) = P(T) = 912mm = P(Kr) + P(O₂) ; Dalton's Law Partial Pressures

Partial Pressure O₂ = P(O₂) = P(T) -P(Kr) = 912mm - 323mm =589mm

Given grams Kr = 8.08g

=> moles Kr = n(Kr) = g(Kr)/fwt(Kr) = 8.08gKr/83.08g/mol = 0.096 mol Kr

Mole Fraction O₂ = X(O₂) = moles O₂ / total moles = n(O₂) / n(ttl)

From PV=nRT => n = PV/RT and substitute into X(O₂) expression gives:

X(O₂) = [P(O₂)·V/n·T] / P(Kr)V/nT + P(O₂)V/nT; 'V/nT cancels'

=> X(O₂) = P(O₂) / P(Kr) + P(O₂) = 589mm/912mm = 0.646

Since X(O₂) = n(O₂) / n(Kr) + n(O₂) = 0.646 = n(O₂) / 0.096molKr + n(O₂)

=> n(O₂) = 0.646[(0.096 + n(O₂)] = 0.062 + 0.646(n(O₂)

=> n(O₂) - 0.646·(n(O₂) = n(O₂)[1 - 0.646] = n(O₂)·0.354 = 0.062

=> n(O₂) = 0.062/0.354 = 0.175 mole O₂

∴ grams O₂ = 0.175 mole O₂ x 32 g/mol O₂ = 5.60 grams O₂