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Dalton's Law of Partial Pressures

10 moles of a...

Question

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

A

8 g

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B

12 g

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C

3 g

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D

6 g

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Solution

The correct option is B 12 g
$n_{A} + n_{B} + n_{C} = 10 moles$
$P_{A} + P_{B} + P_{C} = 10 atm$
also, $P_{A} = 3 atm$ and, $P_{B} = 1 atm$
$∴ P_{C} = 10 - (3 + 1) = 6 atm$
$∴ \frac{n_{C}}{10} \times 10 atm = 6 atm (Dalton's law)$
$∴ n_{C} = 6 ∴ ω_{C} = 6 mol \times 2 g/mol = 12 g$

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