# If a gas effuses 1.89 times as fast as "COCl"_2(g), what is its molar mass?

May 4, 2015

${M}_{r} = 27.7$

Graham's law states that :

$R \propto \frac{1}{{\sqrt{M}}_{r}}$

For 2 gases this becomes:

${R}_{1} / {R}_{2} = \sqrt{\frac{{M}_{2}}{{M}_{1}}}$

${M}_{r} \left[C O C {l}_{2}\right] = 99.0 = {M}_{1}$

${R}_{2} = 1.89 \times {R}_{1}$

So:

$\frac{{R}_{1}}{1.89 {R}_{1}} = \sqrt{\frac{{M}_{2}}{99.0}}$

So:

$\frac{1}{1.89} = \sqrt{\frac{{M}_{2}}{99.0}}$

$0.53 = \sqrt{\frac{{M}_{2}}{99.0}}$

$0.28 = \frac{{M}_{2}}{99.0}$

${M}_{2} = 0.28 \times 99.0 = 27.7$

May 4, 2015

The compound that effuses faster must have lighter molecules, if they are both at the same temperature.
Take the ratio that is $189 : 100 = 1.89$ that means the smaller molecules are in the average moving 1.89 times faster than $C O C {l}_{2}$ molecules.

Square that value, ${\left(\frac{V}{v}\right)}^{2} = {V}^{2} / {v}^{2} = {1.89}^{2} = 3.57$.

This is the mass ratio between heavier and lighter molecules, because these have the same mean kinetic energies and root mean squared speeds, as consequence of the equipartition of energy :

$\frac{1}{2} M {v}^{2} = \frac{1}{2} m {V}^{2}$ implies: ${V}^{2} / {v}^{2} = \frac{M}{m} = 3.57 .$

So the lighter molecules have mass $m = \frac{M}{3.57} = \frac{98.9}{3.57} = 27.7$

where 98.9 = C + O + 2Cl = 12.01 + 16.00 + 35.45ยท2 = 98.9.