# The ratio of rates of diffusion of two gases #X# and #Y# is 2:3. The molar mass of #X# is #27#. Find the molar mass of gas #Y#?

##### 1 Answer

#### 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# - theuniversal gas constant, usually given as#0.0821("atm" * "L")/("mol" * "K")#

#T# - theabsolute temperatureof 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, **molar mass** of the gas,

#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 **density** of the gas. Because the density of the gas is proportional to its molar mass, you can say that

#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)|)))#