Question #8c69f
1 Answer
Explanation:
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)rate∝1√MM
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.
)
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) = 3 xx r_(X)rHe=3×rX
Moreover, you know that
r_(He) prop 1/sqrt(M_"M helium")" "rHe∝1√MM helium and" "r_(X) prop 1/sqrt(M_"M gas X") rX∝1√MM 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")rHerX=1√MM helium⋅√MM 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
Rearrange the above equation to solve for
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")
