Calculate density of a gaseous mixture which consist of 3.01 × 1 0 24 molecules of N 2 and 32 g of O 2 gas at 3 atm pressure and 860 K temperature. (Given : R = 1/12 atm L / mole K, N A = 6.02 × 1 0 23 )

Avg. molar mass = 6 28 × 5 + 32 × 1 = 6 172 From PM = ρ RT = 3 × 6 172 = ρ × 12 1 × 860 ⇒ ρ = 1.2 gm / L

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Calculate density of a gaseous mixture which consist of $3.01 \times 1 0^{24}$ molecules of $N_{2}$ and 32 g of $O_{2}$ gas at 3 atm pressure and 860 K temperature. (Given : $R = 1/12 atm L /$ mole K, $N_{A} = 6.02 \times 1 0^{23}$ )

Options:

Answer:

Solution:

Avg. molar mass $= \frac{28 \times 5 + 32 \times 1}{6} = \frac{172}{6}$

From $PM = ρ RT$

$= 3 \times \frac{172}{6} = ρ \times \frac{1}{12} \times 860$

$\Rightarrow ρ = 1.2 gm / L$

Stream:JEESubject:ChemistryTopic:Ideal GasSubtopic:Ideal Gas Equation & problems

⏱ 2mℹ️ Source: QB

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