Respuesta :
Answer:
For (a): The moles of Ar is 16.94 moles
For (b): The moles of [tex]Cl_2[/tex] is 16.94 moles
For (c): The total number of moles in a tank is 23.47 moles
For (d): The mole fraction of Ar is 0.722
For (e): The mole fraction of [tex]Cl_2[/tex] is 0.278
For (f): The partial pressure of Ar is 2.888 atm
For (g): The partial pressure of [tex]Cl_2[/tex] is 1.112 atm
Explanation:
The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)
- For (a):
Given mass of Ar = 675.5 g
Molar mass of Ar = 39.95 g/mol
Plugging values in equation 1:
[tex]\text{Moles of Ar}=\frac{675.5g}{39.95g/mol}=16.91 mol[/tex]
- For (b):
Given mass of [tex]Cl_2[/tex] = 465.0 g
Molar mass of [tex]Cl_2[/tex] = 70.9 g/mol
Plugging values in equation 1:
[tex]\text{Moles of }Cl_2=\frac{465.0g}{70.9g/mol}=6.56 mol[/tex]
- For (c):
Total moles of gas in the tank = [16.91 + 6.56] mol = 23.47 mol
Mole fraction is defined as the moles of a component present in the total moles of a solution. It is given by the equation:
[tex]\chi_A=\frac{n_A}{n_A+n_B}[/tex] .....(2)
where n is the number of moles
- For (d):
Moles of Ar = 16.94 moles
Total moles of gas in the tank = 23.47 mol
Putting values in equation 2, we get:
[tex]\chi_{Ar}=\frac{16.94}{23.47}\\\\\chi_{Ar}=0.722[/tex]
- For (e):
Total mole fraction of the system is always 1
Mole fraction of [tex]Cl_2[/tex] = [1 - 0.722] = 0.278
Raoult's law is the law used to calculate the partial pressure of the individual gases present in the mixture.
The equation for Raoult's law follows:
[tex]p_A=\chi_A\times p_T[/tex] .....(3)
where [tex]p_A[/tex] is the partial pressure of component A in the mixture and [tex]p_T[/tex] is the total partial pressure of the mixture
- For (f):
We are given:
[tex]\chi_{Ar}=0.722\\p_T=4.00atm[/tex]
Putting values in equation 3, we get:
[tex]p_{Ar}=0.722\times 4.00atm\\\\p_{Ar}=2.888atm[/tex]
- For (g):
We are given:
[tex]\chi_{Cl_2}=0.278\\p_T=4.00atm[/tex]
Putting values in equation 3, we get:
[tex]p_{Cl_2}=0.278\times 4.00atm\\\\p_{Cl_2}=1.112atm[/tex]
