<u>Answer:</u> The volume and total pressure of the mixture is 0.876 L and 89.36 kPa respectively.
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
.....(1)
- <u>For methane:</u>
Given mass of methane = 320 mg = 0.3 g (Conversion factor: 1 g = 1000 mg)
Molar mass of methane = 16 g/mol
Putting values in equation 1, we get:
- <u>For argon:</u>
Given mass of argon = 175 mg = 0.175 g
Molar mass of argon = 40 g/mol
Putting values in equation 1, we get:
- <u>For nitrogen:</u>
Given mass of nitrogen = 225 mg = 0.225 g
Molar mass of nitrogen = 28 g/mol
Putting values in equation 1, we get:
To calculate the volume of the mixture, we use the equation:
PV = nRT ......(2)
We are given:
Partial pressure of argon = 12.52 kPa
Temperature = 300 K
R = Gas constant =
n = number of moles of argon = 0.0044 moles
Putting values in equation 2, we get:
Now, calculating the total pressure of the mixture by using equation 2:
Total number of moles = [0.019 + 0.0044 + 0.0080] mol = 0.0314 mol
V= volume of the mixture = 0.876 L
Putting values in equation 2, we get:
Hence, the volume and total pressure of the mixture is 0.876 L and 89.36 kPa respectively.