# Question 6cee9

Apr 18, 2015

The ratio of rates $C {l}_{2} : {N}_{2} = 0.63$

Graham's Law states that:

Rate of effusion $\propto \frac{1}{\sqrt{{M}_{r}}}$

Where ${M}_{r}$ is the relative molecular mass.

For two gases:

R_1/R_2=sqrt(M_(r(2))/(M_(r(1)))

${M}_{r} \left[C {l}_{2}\right] = \left(35.5 \times 2\right) = 71$

${M}_{r} \left[{N}_{2}\right] = \left(14 \times 2\right) = 28$

So R_(Cl_2)/(R_(N_2)$= \sqrt{\frac{28}{71}} = 0.63$

Apr 18, 2015

Use Grahams's law, which states that the rate of effusion of a gas is inversely proportional to the square root of molar mass.

${r}_{\text{eff}} \propto \frac{1}{\sqrt{{M}_{M}}}$

This means that the rate of effusion of chlorine gas will be

r_"chlorine" prop 1/sqrt(M_("M chlorine"))

Likewise, the rate of effusion of nitrogen gas will be

r_"nitrogen" prop 1/sqrt(M_("M nitrogen"))

If you divide these two expressions, you'll get the ratio of effusion rates of chlorine and nitrogen

r_"chlorine"/r_"nitrogen" = (1/sqrt(M_("M chlorine")))/(1/sqrt(M_("M nitrogen"))) = sqrt(M_("M nitrogen"))/sqrt(M_("M chlorine"))

The numerical value of this ratio will be

r_"chlorine"/r_"nitrogen" = (sqrt(28.014cancel("g/mol")))/(sqrt(70.906cancel("g/mol"))) = color(green)(0.629)#

Here is a video which shows how to solve a different problem using Graham's law.

Video from: Noel Pauller

Hope this helps!