Answer:
a) 24.31 g
b) 16.69 g
Explanation:
A mixture of CO2 and Kr weighs 41.0 g and exerts a pressure of 0.729 atm in its container.
After the CO2 is completely removed by absorption with NaOH(s), the pressure in the container is 0.193 atm.
Therefore, Pressure of Kr = 0.193 atm
Pressure of CO2 = 0.729 - 0.193 = 0.536 atm
Their mole fraction can be also determined as follows:
CO2 = [tex]\frac{Presssure due to CO_2}{Total pressure}[/tex]
CO2 = [tex]\frac{0.536}{0.729}[/tex]
= 0.735
Also; for Kr ; we have
Kr = [tex]\frac{0.193}{0.729}[/tex]
Kr = 0.265
Molar mass of CO2 = 44 g/mol
Molar mass of Kr = 83.78 g/mol
Mass of CO2 = mole fraction * molar mass = 0.735 * 44 = 32.34
Mass of Kr = 0.265 * 83.78 = 22.20
Total mass = 32.34 +22.20 = 54.54
The Percentage of gas in mixture is as follows:
% CO2 = [tex]\frac{32.34}{54.54}[/tex]* 100 %
= 0.5930
= 59.30%
(a) Mass of CO2 in mixture = 0.5930* 41 g = 24.31 g
% Kr =[tex]\frac{22.20}{ 54.54}[/tex] * 100 %
= 0.407
= 40.70 %
(b) Mass of Kr in mixture = 0.407 * 41 = 16.69 g
