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At 50.0 oC, a reinforced tank contains 675.5 grams of gaseous argon and 465.0 g of gaseous molecular chlorine with a total pressure of 4.00 atm. Calculate the following:
a. How many moles of Ar are in the tank?
b. How many moles of Cl, are in the tank?
c. Total moles of gas in the tank.
d. The mole fraction of Ar.
e. The mole fraction of Cl2.
f. The Partial Pressure of Ar.
g. The Partial Pressure of Cl2.

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]

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