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3 years ago

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A mixture of 1.20 mols He, 2.40 mols Ne, 4.80 mols Kr, and 0.60 mols Ar has a total pressure of 600.0 mm Hg. What is the partial

pressure of the Kr? Show all work to receive full credit.

1 answer:

Juliette [100K]3 years ago

8 0

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

$p_A=x_A\times P_{total}$

where, $p_A$ = partial pressure of component A

$x_A$ = mole fraction of A

$P_{total}$ = total pressure

mole fraction of Krypton = $\frac{\text {Moles of Kr}}{\text {total moles}}=\frac{4.80}{1.20+2.40+4.80+0.60}=0.53$

$p_{Kr}=0.53\times 600mmHg=320mmHg$

Thus partial pressure of the Kr is 320 mm Hg

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