# 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?

Jun 8, 2016

${\text{12 g mol}}^{- 1}$

#### 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

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} P V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$, 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 as $0.0821 \left(\text{atm" * "L")/("mol" * "K}\right)$
$T$ - 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, $m$, and the molar mass of the gas, ${M}_{M}$

$n = \frac{m}{M} _ M$

you can say that

$P V = \frac{m}{M} _ M \cdot R T$

Rearrange this to get

$P \cdot {M}_{M} = {\overbrace{\frac{m}{V}}}^{\textcolor{b l u e}{\setminus \rho}} \cdot R T$

$\rho = \frac{P}{R T} \cdot {M}_{M}$

Here $\rho$ represents the density of the gas. Because the density of the gas is proportional to its molar mass, you can say that

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{rate of diffusion} \propto \frac{1}{\sqrt{{M}_{M}}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

So, you know that for two gases $X$ and $Y$, you have

${\text{rate"_ (X)/"rate}}_{Y} = \frac{2}{3}$

Since

$\text{rate"_ X = 1/sqrt(M_"M X")" }$ 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 ${M}_{\text{MY}}$

${M}_{\text{M Y" = ("rate"_x/"rate"_Y)^2 * M_"M X}}$

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)|)))