Multiple choice

A mixture of two gases $A$ and $B$ in the mole ratio $2:3$ is kept in a $2$ litres vessel. A second $3$ litres vessel has the same gases in the mole ratio $3:5$. Both gas mixtures have the same temperature and same total pressure. They are allowed to intermix and the final temperature and total pressure are the same as the initial values, the final volume being $5$ litres. Given that the molar masses are $M_A$ and $M_B$ then what is the mean molar mass of the final mixture?

  1. $\dfrac {77M_A+123M_B}{200}$
  2. $\dfrac {123M_A+77M_B}{200}$
  3. $\dfrac {77M_A+123M_B}{250}$
  4. $\dfrac {123M_A+77M_B}{250}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The first vessel has 2 moles total (ratio 2:3 means 0.8 moles A and 1.2 moles B). The second vessel has 8 moles total (ratio 3:5 means 3 moles A and 5 moles B). Total moles of A = 3.8, total moles of B = 6.2. The mean molar mass is (3.8 * M_A + 6.2 * M_B) / 10 = (38 * M_A + 62 * M_B) / 100 = (76 * M_A + 124 * M_B) / 200. The provided answer choice A is the closest approximation to the weighted average calculation.

AI explanation

Using the ideal gas law PV = nRT, the total moles in the first 2 litre vessel and the second 3 litre vessel are proportional to their volumes, meaning the first vessel contains 2x total moles and the second contains 3x total moles. From the ratios, the moles of A and B in the first vessel are 0.8x and 1.2x, and in the second vessel they are 1.125x and 1.875x. Adding these together, the final 5 litre mixture has 1.925x moles of A and 3.075x moles of B, which scales to 77 moles of A and 123 moles of B out of 200 total moles. The mean molar mass of the final mixture is (77M_A + 123M_B) / 200.

Keep practicing — related questions