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 $2$ g/mol. Then the weight of $C$ present in the mixture will be:

8

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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, $n_{A} = 3$ , $n_{B} = 1$
$∵ P_{A} = x_{A} \times P_{t o t a l} = \frac{n_{A}}{n_{A} + n_{B} + n_{C}} \times 10 = \frac{n_{A}}{10} \times 10 = 3$
Similarly,
$P_{B} = x_{B} \times P_{t o t a l}$
So $n_{B} = 1$
$∴ n_{C} = 10 - (3 + 1) = 6$
Weight of $C = 6 \times 2 = 12$ g

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