CO(g) effuses at a rate that is ______ times that of Xe(g) under the same conditions.

asked by Alex
  1. 2.16

    posted by Alex
  2. Let's make up a number for the rate of Xe gas. Let's call it 10 mL/second.

    (rate CO/rate Xe) = sqrt(M Xe/M CO where M stands for molar mass Xe or molar mass CO.

    (10/x) = sqrt(131.29/28)
    Solve for x for rate of CO then calculate how much faster/slower this is than Xe. You know CO should have a higher rate since the molar mass is lower.

    posted by DrBob222

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