Question #8c69f

The idea here is that the rate of effusion of a gas, that is, the number of moles of gas per unit of time, is inversely proportional to the square root of its molar mass - this is known as Graham's Law.

#"rate" prop 1/sqrt(M_M)#

In other words, the heavier each individual molecule of a gas is, the slower it will effuse. LIkewise, the lighter the molecules of a gas, the faster they will effuse.

Here's how that would look for helium and ethylene oxide - notice that fewer molecules of ethylene oxide effuse when compared with the lighter helium molecules.

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In your case, you know that a certain gas takes three times as long to effuse as jelium. Right from the start, you know that you're dealing with a gas that has heavier molecules than helium does.

If you take #r_(He)# to be the rate of effusion of helium, and #r_(X)# to be the rate of effusion of the unknown gas, you can write

#r_(He) = 3 xx r_(X)#

Moreover, you know that

#r_(He) prop 1/sqrt(M_"M helium")" "# and #" "r_(X) prop 1/sqrt(M_"M gas X")#

If you divide these expressions, you will find that

#r_(He)/r_(X) = 1/sqrt(M_"M helium") * sqrt(M_"M gas X")#

#(3 xx color(red)(cancel(color(black)(r_(X)))))/(color(red)(cancel(color(black)(r_(X))))) = sqrt(M_"M gas X")/sqrt(M_"M helium")#

To keep the calculations simple, you can use helium's molar mass to be #"4 g/mol"#.

Rearrange the above equation to solve for #M_"M gas X"#

#sqrt(M_"M gas X") = 3 xx sqrt(M_"M helium")#

Square both sides to get rid of the square roots

#(sqrt(M_"M gas X"))^2 = 3^2 xx (sqrt(M_"M helium"))^2#

#M_"M gas X" = 9 xx "4 g/mol" = color(green)("36 g/mol")#