Multiple choice

The total weight of a gas mixture which contains $11.2$ litres of $CO_{2}$, $5.6$ litres of $O_{2}$, $22.4$ litres of $N_{2}$, $44.8$ litres of $He$ at $0^{o}C$ and $760$ $mm$ of $Hg$ pressure is:

  1. $44$ $g$
  2. $22$ $g$
  3. $33$ $g$
  4. $66$ $g$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

At STP, 22.4 L = 1 mole. Moles: CO2 = 0.5, O2 = 0.25, N2 = 1, He = 2. Masses: CO2 = 0.5 * 44 = 22g, O2 = 0.25 * 32 = 8g, N2 = 1 * 28 = 28g, He = 2 * 4 = 8g. Total mass = 22 + 8 + 28 + 8 = 66g.

AI explanation

At STP, 1 mole of any ideal gas occupies 22.4 liters, so dividing each gas volume by 22.4 yields the respective moles. The mixture contains 0.5 moles of CO2 (22 g), 0.25 moles of O2 (8 g), 1 mole of N2 (28 g), and 2 moles of He (8 g). Adding these individual masses together gives a total weight of 22 + 8 + 28 + 8 = 66 g.

Keep practicing — related questions