An enclosure of volume $V$ contains a mixture of $8 g$ of oxygen, $14 g$ of nitrogen and $22 g$ of carbon- dioxide at absolute temperature $T$ . The pressure of the mixture of gases is ( $R$ is universal gas constant)

Join BYJU'S Learning Program

The average molecular mass of a mixture of gas containing nitrogen and carbon dioxide is 36. The mixture contains 280 g of nitrogen. Therefore, the amount of CO2 present in the mixture is :

The correct option is C 440 gMoles of N2=28028=10moles Assume, moles of CO2=x. So, average molecular mass of mixture =10×28+x×4410+x =36 (given)So, x=10 molesTherefore, weight of CO2=10×44=440 g

The average molecular mass of a mixture of gas containing nitrogen and carbon dioxide is 36. The mixture contains 280 g of nitrogen. Therefore, the amount of ${C O}_{2}$ present in the mixture is :

The correct option is C 440 g
Moles of $N_{2} = \frac{280}{28} = 10 m o l e s$
Assume, moles of $C O_{2} = x$ .
So, average molecular mass of mixture $= \frac{10 \times 28 + x \times 44}{10 + x}$ $= 36$ (given)
So, $x = 10$ moles
Therefore, weight of $C O_{2} = 10 \times 44 = 440$ g