Respuesta :
Explanation:
Since, it is given that carbon dioxide is completely removed by absorption with NaOH. And, pressure inside the container is 0.250 atm.
For Kr = 0.250 atm and pressure [tex]CO_{2}[/tex] will be calculated as follows.
[tex]CO_{2}[/tex] = (0.708 - 0.250) atm
= 0.458 atm
Now, we will calculate the mole fraction as follows.
[tex]CO_{2} = \frac{0.458}{0.708}[/tex]
= 0.646
Kr = [tex]\frac{0.250}{0.708}[/tex]
= 0.353
Now, we will convert into gram fraction as follows.
[tex]CO_{2} = 0.646 \times 44[/tex]
= 28.424
Kr = [tex]0.353 \times 83.78[/tex]
= 29.57
Therefore, total mass is calculated as follows.
Total mass = (28.424 + 29.57)
= 57.994
Hence, the percentage of [tex]CO_{2}[/tex] and Kr are calculated as follows.
[tex]CO_{2} = \frac{28.424}{57.99} \times 100[/tex]
= 49%
Kr = [tex]\frac{29.57}{57.99} \times 100[/tex]
= 51%
Hence, amount of [tex]CO_{2}[/tex] and Kr present i mixture is as follows.
[tex]CO_{2}[/tex] in mixture = [tex]35 \times 0.49[/tex]
= 17.15 g
Kr = [tex]35 \times 0.51[/tex]
= 17.85 g
Thus, we can conclude that 17.15 g of [tex]CO_{2}[/tex] is originally present and 17.85 g of Kr is recovered.
