A gasesous mixture contains three gases $N_{2}$ , $O_{2}$ and $C O_{2}$ with a total number of moles of $10$ and total pressure of $10 atm$ . The partial pressure of $N_{2}$ is $3 atm$ . The pressure is reduced to $9 a t m$ if the gases is passed through alkaline pyrogallol. If $C O_{2}$ has molecular weight of $44 g/mol$ . Then, the weight of $C O_{2}$ (in grams) present in the mixture will be:

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Solution

Given total pressure $(P) = 10 atm$
Given,
total number of moles; $n_{N_{2}} + n_{O_{2}} + n_{C O_{2}} = 10$
The gas $O_{2}$ is absorbed by pyrogallol. Hence the loss of pressure is due to $O_{2}$ .

$10 - \frac{n_{O_{2}}}{n_{N_{2}} + n_{O_{2}} + n_{C O_{2}}} \times 10 = 9 a t m$

Hence the number of moles of $O_{2}$ is 1.
$P_{N_{2}} = 3 atm, P_{O_{2}} = 1 atm, n_{N_{2}} = 3, n_{O_{2}} = 1$
$∵ P_{N_{2}} = x_{N_{2}} \times P_{(total)} = \frac{n_{N_{2}}}{n_{N_{2}} + n_{O_{2}} + n_{C O_{2}}} \times 10$
$= \frac{n_{N_{2}}}{10} \times 10 \Rightarrow n_{N_{2}} = 3$

$∴ n_{C O_{2}} = 10 - (n_{N_{2}} + n_{O_{2}}) = 10 - 4 = 6$
Weight of $C = 6 \times 44 = 264 g$

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