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Answer: The partial pressure of the Kr is 320 mm Hg.
Explanation:
According to Raoult's Law , the partial pressure of each component in the solution is equal to the total pressure multiplied by its mole fraction. It is mathematically expressed as
[tex]p_A=x_A\times P_{total}[/tex]
where, [tex]p_A[/tex] = partial pressure of component A
[tex]x_A[/tex] = mole fraction of A
[tex]P_{total}[/tex] = total pressure
mole fraction of Krypton = [tex]\frac{\text {Moles of Kr}}{\text {total moles}}=\frac{4.80}{1.20+2.40+4.80+0.60}=0.53[/tex]
[tex]p_{Kr}=0.53\times 600mmHg=320mmHg[/tex]
Thus partial pressure of the Kr is 320 mm Hg
The partial pressure of krypton in the given mixture if total pressure is 600 mmHg is 320mmHg.
How do we calculate partial pressure?
Partial pressure of any gas will be calculated by using the Rault's Law as:
p = (x)(P), where
p = partial pressure
P = total pressure = 600 mmHg
x = mole fraction
Mole fraction of krypton = moles of krypton/total moles
x = 4.80 / (1.20+2.40+4.80+0.60)
x = 0.53
Partial pressure of krypton will be calculated as:
p = (0.53)(600) = 320mmHg
Hence partial pressure of Kr is 320mmHg.
To know more about Rault's law, visit the below link:
https://brainly.com/question/10165688
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