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 and if C has molecular weight of 2g/mol. Then,the weight of C present in the mixture will be:

8 g

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12 g

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3 g

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6 g

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Solution

The correct option is D 12 g
Given that,

$P_{T o t a l} = 10 a t m$ , $η_{T} = 10$

$P_{C} = P_{T} - (P_{A} + P_{B}) = 10 - 4 = 6$

$∴ \frac{η_{C}}{η_{T}} = \frac{P_{C}}{P_{T}}$

$\frac{W_{C}}{M_{W}} = \frac{P_{C} \times η_{T}}{P_{T}} \Rightarrow W_{C} = \frac{6 \times 10 \times 2}{10}$

$= 12 g$

So, the correct option is $B$

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