Question

If percentage of N$_2$ above a liquid solution is 80% at a total pressure of 10 atm, then find the mole fraction of N$_2$ gas dissolved in solution.
[Given that Henry’s constant for N$_2$ is $7.6 \times 10^7$ mm Hg]

• A. $10^{-4}$
• B. $8 \times 10^{-5}$
• C. $10^{-7}$
• D. $10^{-6}$

Hint:

Always convert pressure into the same unit as Henry’s constant before substitution.

Answer 1

The Correct Option is B

Step 1: Apply Henry’s law

According to Henry’s law:

P = k_H · X

where
P = partial pressure of the gas,
k_H = Henry’s constant,
X = mole fraction of the gas in solution.

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Step 2: Calculate partial pressure of N₂

Total pressure = 10 atm
Percentage of N₂ = 80%

P_N₂ = 0.8 × 10 = 8 atm

Convert atm to mm Hg:

P_N₂ = 8 × 760 = 6080 mm Hg

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Step 3: Substitute values in Henry’s law

Henry’s constant for N₂:

k_H = 7.6 × 10⁷ mm Hg

6080 = 7.6 × 10⁷ · X

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Step 4: Solve for mole fraction

X = 6080 / (7.6 × 10⁷)

X = 8 × 10⁻⁵

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Final Answer:

The mole fraction of N₂ gas dissolved in the solution is
8 × 10⁻⁵