Answer : The mass of helium (He) in this container is, 0.2 grams
Solution :
First we have to calculate the moles of gas present in the container.
Using ideal gas equation,
[tex]PV=nRT\\\\n=\frac{RT}{PV}[/tex]
where,
P = pressure of the gas present in the container = 1.40 atm
V = volume of the gas present in the container = 226.4 L
R = gas constant = 0.0821 Latm/moleK
T = temperature of the gas = [tex]27^oC=273+27=300K[/tex]
n = moles of gas present in the container = ?
Now put all the given values in the above equation, we get the moles of gas present in the container.
[tex]n=\frac{(0.0821Latm/moleK)\times (300K)}{(226.4L)\times (1.40atm)}=0.077moles[/tex]
Now we have to calculate the moles of helium.
[tex]\text{Moles of helium}=0.077\times \frac{65.5}{100}=0.050mole[/tex]
Now we have to calculate the mass of helium present in the container.
[tex]\text{Mass of helium}=\text{Moles of helium}\times \text{Molar mass of helium}=0.050moles\times 4g/mole=0.2g[/tex]
Therefore, the mass of helium (He) in this container is, 0.2 grams
