Multiple choice

A mixture of ${CH}{4}$ and ${O}{2}$ have vapour density $12$ then % of ${CH}_{4}$, by volume in that mixture is:

  1. 40%

  2. 60%

  3. 50%

  4. 75%
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Vapour density = M_mix / 2 = 12, so M_mix = 24. M_mix = (x * 16 + (1-x) * 32) = 24. 16x + 32 - 32x = 24. 16x = 8. x = 0.5 = 50%.

AI explanation

Using the mixture vapour density formula, 12 = (16x + 32(1-x)) / 2, where x is the mole fraction of methane. Solving this equation yields a mole fraction of 0.5 for methane, which equals a 50% volume percentage since the volume fraction of ideal gases equals their mole fraction.

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