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A neon-dioxygen mixture contains 70.6 g O2 and 167.5 g neon. If pressure of the mixture of gases in the cylinder is 25 bar.

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A neon-dioxygen mixture contains 70.6 g O2 and 167.5 g neon. If pressure of the mixture of gases in the cylinder is 25 bar. What is the partial pressure of O2 and Ne in the mixture respectively?

(a) 5.25 bar, 10 bar

(b) 19.75 bar, 5.25 bar

(c) 19.75 bar, 10 bar

(d) 5.25 bar, 19.75 bar

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Correct answer: (d) 5.25 bar, 19.75 bar

Explanation:

Number of moles of O2

= \(\frac{70.6g}{32g\;mol^{-1}}\) = 2.21 mole

Number of moles of Ne

= \(\frac{167.5g}{20g\;mol^{-1}}\) = 8.375 mole

Mole fraction of O2

= \(\frac{2.21}{2.21 + 8.375}\) = 0.21 mole

Mole fraction of Ne = 1 – 0.21 = 0.79

Partial pressure of a gas = Mole fraction × total pressure

Partial pressure of O2 = 0.21 × 25 = 5.25 bar

Partial pressure of Ne = 0.79 × 25 = 19.75 bar

This post was modified 3 years ago by Shivani siva
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