Answer:
The mixture contains 8.23 g of Ar
Explanation:
Let's solve this with the Ideal Gases Law
Total pressure of a mixture = (Total moles . R . T) / V
We convert T° from °C to K → 85°C + 273 = 358K
3.43 atm = (Moles . 0.082 L.atm/mol.K . 358K) / 6.47L
(3.43 atm . 6.47L) / (0.082 L.atm/mol.K . 358K) = Moles
0.756= Total moles from the mixture
Moles of Ar + Moles of H₂ = 0.756 moles
Moles of Ar + 1.10 g / 2g/mol = 0.756 moles
Moles of Ar = 0.756 moles - 0.55 moles H₂ → 0.206
We convert the moles to g → 0.206 mol . 39.95 g / 1 mol = 8.23 g
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