Multiple choice

A mixture of $Ne$ and $Ar$ at $250$K has a total K.E. $=3$ KJ in a closed vessel, the total mass of $Ne$ and $Ar$ is $30$g. Find mass $\%$ of $Ne$ in the gaseous mixture at $250$K.

  1. $61.63$
  2. $38.37$
  3. $50$
  4. $28.3$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Total Kinetic Energy = 3/2 * n * R * T. With T=250K, 3000 = 1.5 * n * 8.314 * 250, so n = 3000 / 3117.75 = 0.9622 moles. Let m be mass of Ne, then (m/20) + ((30-m)/40) = 0.9622. Multiplying by 40: 2m + 30 - m = 38.488, so m = 8.488g. Mass % = (8.488/30)*100 = 28.29%.

Keep practicing — related questions