Multiple choice

A synthetic mixture of nitrogen and Argon has a density of $1.4 g L^{-1}$ at $0^oC$. Find out the volume percentage of nitrogen in the mixture?

  1. $85$
  2. $96$
  3. $72$
  4. $60$
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C Correct answer
Explanation

Density = PM / RT. M_avg = dRT / P. At 0C (273K) and 1 atm, M_avg = 1.4 * 0.0821 * 273 / 1 = 31.37 g/mol. Let x be mole fraction of N2 (28 g/mol), 1-x be Ar (40 g/mol). 28x + 40(1-x) = 31.37. 12x = 8.63. x = 0.719 = 72%.

AI explanation

Let the volume percentage of nitrogen be x, making the volume percentage of argon (100 minus x). Using the formula for the density of a gas mixture, 1.4 equals the mass of the gases divided by 22.4 liters, so the total mass is 31.36 grams. Setting up the equation (28x plus 40(100 minus x)) divided by 100 equals 31.36, we solve for x to find the nitrogen percentage. This calculation yields a nitrogen volume percentage of approximately 72.

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