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A gaseous mixture contains three gases A, B and C with a total number of moles of 10 and total pressure of 10 atm. The partial pressure of A and B are 3 atm and 1 atm respectively. If C has molecular weight of 2 g/mol, then the weight of C present in the mixture will be

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8 g

12 g

3 g

6 g

answer is B.

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Detailed Solution

$P_{total} = 10 atm, n_{T} = 10 P_{C} = P_{T} - (P_{A} + P_{B}) = 10 - 4 = 6 ∴ \frac{n_{C}}{n_{T}} = \frac{P_{C}}{P_{T}} \frac{W_{C}}{M_{W}} = \frac{P_{C} n_{T}}{P_{T}} \Rightarrow W_{C} = \frac{6 \times 10 \times 2}{10} = 12 g$

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