The ratio of rates of diffusion of two gases XX and YY is 2:3. The molar mass of XX is 2727. Find the molar mass of gas YY?
1 Answer
Explanation:
An important thing to keep in mind here is that you're dealing with rates of diffusion, not with rates of effusion. Because the two gases are presumably kept under the same conditions for pressure and temperature, you can find a relationship between their rates of diffusion and their molar masses.
Now, according to Graham's Law of Diffusion, the rate of diffusion of a gas is inversely proportional to the square root of its density
color(blue)(|bar(ul(color(white)(a/a)"rate" prop 1/sqrt("density")color(white)(a/a)|)))
However, because the density of a gas is directly proportional to its molar mass, you can use the ideal gas law equation
color(blue)(|bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" " , where
P - the pressure of the gas
V - the volume it occupies
n - the number of moles of gas
R - the universal gas constant, usually given as0.0821("atm" * "L")/("mol" * "K")
T - the absolute temperature of the gas
to show that the rate of diffusion of a gas is also inversely proportional to the square root of molar mass.
Since you know that the number of moles can be expressed as the ratio between the mass of the sample,
n = m/M_M
you can say that
PV = m/M_M * RT
Rearrange this to get
P * M_M = overbrace(m/V)^(color(blue)(\rho)) * RT
rho = P/(RT) * M_M
Here
color(purple)(|bar(ul(color(white)(a/a)color(black)("rate of diffusion" prop 1/sqrt(M_M))color(white)(a/a)|)))
So, you know that for two gases
"rate"_ (X)/"rate"_(Y) = 2/3
Since
"rate"_ X = 1/sqrt(M_"M X")" " and" " "rate"_ Y = 1/sqrt(M_"M Y")
you can say that you have
"rate"_ X/"rate"_ Y = 1/sqrt(M_"M X") * sqrt(M_"M Y") = sqrt(M_"M Y"/M_"M X")
Square both sides of the equation and rearrange to solve for
M_"M Y" = ("rate"_x/"rate"_Y)^2 * M_"M X"
Plug in your values to find
M_"M Y" = (2/3)^2 * "27 g mol"^(-1) = color(green)(|bar(ul(color(white)(a/a)color(black)("12 g mol"^(-1))color(white)(a/a)|)))
