# Question #21958

##### 1 Answer

#### Answer:

#### Explanation:

The idea here is that you need to use **Graham's Law of Diffusion** to find a relationship between the *volumes* of the two gases and their **molar masses**.

When two gases are kept under the same conditions for *pressure* and *temperature*, the ratio that exists between their number of moles is **equivalent** to the ratio that exists between their **volumes**.

You can prove this by using the **ideal gas law** equation

#color(blue)(|bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)|)))#

Here

*universal gas constant*, usually given as

**absolute temperature** of the gas

You can write

#P * V_A = n_A * RT -># for gas#"A"#

#P * V_B = n_B * RT -># for gas#"B"#

Divide these two equations to get

#(color(red)(cancel(color(black)(P))) * V_A)/(color(red)(cancel(color(black)(P))) * V_B) = (n_A * color(red)(cancel(color(black)(RT))))/(n_B * color(red)(cancel(color(black)(RT)))) implies color(purple)(|bar(ul(color(white)(a/a)color(black)(V_A/V_B = n_A/n_B)color(white)(a/a)|)))#

This means that the diffusion of a volume of gas is equivalent to the diffusion of a *number of moles* of gas.

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)|)))#

You can use the ideal gas law equation to write the density of a gas as a function of its **molar mass**. Keep in mind that you have

#n = m/M_M#

Here *mass* and **molar mass**. You will thus have

#PV = m/M_MRT#

Rearrange to get

#P * M_M = m/V * RT#

#P/(RT) * M_M = overbrace(m/V)^(color(blue)("density"))#

So, the rate of diffusion can be expressed in terms of the **volumes** of the two gases and of their **molar masses**

#"rate"_A prop 1/sqrt(P/(RT) * M_"M A")" "# and#" " "rate"_B prop 1/sqrt(P/(RT) * M_"M B")#

You can say that if **same period of time** as

#"rate"_A/"rate"_B = (50 color(red)(cancel(color(black)("mL"))))/(40color(red)(cancel(color(black)("mL")))) = 5/4#

This is equivalent to

#5/4 = sqrt(P/(RT) * M_"M B")/sqrt(P/(RT) * M_"M A") = color(red)(cancel(color(black)(sqrt(P/(RT)))))/color(red)(cancel(color(black)(sqrt(P/(RT))))) * sqrt(M_"M B")/sqrt(M_"M A")#

Since you know that

#M_"M A" = "64 g mol"^(-1)#

you can say that

#5/4 = sqrt(M_"M B")/sqrt(64)#

This will get you

#sqrt(M_"M B") = 5/4 * sqrt(64) = 10#

Therefore,

#M_"M B" = 10^2 = color(green)(|bar(ul(color(white)(a/a)"100 g mol"^(-1)color(white)(a/a)|)))#