If I takes 4.67 times as long for a particular gas to effuse as it takes hydrogen under the same conditions, what is the molecular mass of the gas? (Note that the rate of effusion is inversely proportional to the time it takes for a gas to effuse.)
Hi, If you find section in your textbook for Graham's Law of Effusion (or look it up online, e.g., here: http://klchem.weebly.com/), you can plug in the molar mass of hydrogen gas and leave the other molar mass as the variable. Then plug in X for the rate of effusion for H2 and 4.67X for the rate of effusion of the other gas. The X's cancel out, so you can solve for the unknown molar mass. Make sure you keep the 1's and 2's straight in the formula. Remember, the larger molecules are slower, so you can check your answer to see if it makes sense. Good luck!