A gaseous mixture contains three non-reacting 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 $2 {g mol}^{- 1}$ . 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 B $12 g$
Given,
$P = 10 atm$ ,
Total numbers of moles; $n_{A} + n_{B} + n_{C} = 10$
$P_{A} = 3 atm, P_{B} = 1 atm$
$∵ P_{A} = x_{A} \times P_{(t o t a l)} = \frac{n_{A}}{n_{A} + n_{B} + n_{C}} \times 10 = \frac{n_{A}}{10} \times 10 = n_{A} ∴ P_{A} = n_{A} = 3 mol$
Similarly, $P_{B} = x_{B} \times P_{(t o t a l)}$
So, $n_{B} = 1 mol$

$∴ n_{C} = 10 - (n_{A} + n_{B}) = 10 - 4 = 6$
$Weight of C = 6 \times 2 = 12 g$

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