Answer:
The partial pressure will be "100 torr".
Explanation:
Given:
[tex]P_{Ar} = 300 \ torr[/tex]
By assuming Ar and Ne having 50 gm each, we get
mol of Ne = [tex]\frac{50}{20}[/tex]
= [tex]2.5 \ mol[/tex]
mol of Ar = [tex]\frac{50}{40}[/tex]
= [tex]1.25 \ mol[/tex]
now,
[tex]n_T= mol.A_r+mol.N_e[/tex]
[tex]=1.25+2.5[/tex]
[tex]=3.75[/tex]
then,
[tex]X_{Ar}=\frac{n_{Ar}}{n_T}[/tex]
[tex]=\frac{1.25}{3.75}[/tex]
[tex]=0.33[/tex]
hence,
The partial pressure of Ar will be:
⇒ [tex]P_{Ar} = P_T\times X_{AT}[/tex]
By substituting the values, we get
[tex]=300\times 0.33[/tex]
[tex]=100 \ torr[/tex]
The partial pressure of Ar in the mixture is 99.9 torr
Let the mass of both gas be 10 g
Next, we shall determine mole of each gas.
Mass = 10 g
Molar mass of Ne = 20 g/mol
Mole = mass / molar mass
Mole of Ne = 10 / 20
Mass = 10 g
Molar mass of Ar = 40 g/mol
Mole = mass / molar mass
Mole of Ar = 10 / 40
Next, we shall determine the mole fraction of Ar
Mole of Ne = 0.5 mole
Mole of Ar = 0.25 mole
Total mole = 0.5 + 0.25 = 0.75 mole
[tex]mole \: fraction \: = \frac{mole}{total \: mole} \\ \\ mole \: fraction \: of \:Ar = \frac{0.25}{0.75} \\ \\ mole \: fraction \: of \:Ar = 0.333 \\ \\ [/tex]
Finally, we shall determine the partial pressure of Ar
Mole fraction of Ar = 0.333
Total pressure = 300 torr
Partial pressure = mole fraction × total pressure
Partial pressure of Ar = 0.333 × 300
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