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Question #6cee9

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Answer 1 · 2015-04-18T17:45:04

The ratio of rates $C l_{2} : N_{2} = 0.63$ $Cl_2:N_2=0.63$

Graham's Law states that:

Rate of effusion $\propto \frac{1}{\sqrt{M_{r}}}$ $prop1/sqrt(M_r)$

Where $M_{r}$ $M_r$ is the relative molecular mass.

For two gases:

$\frac{R_{1}}{R_{2}} = \sqrt{\frac{M_{r (2)}}{M_{r (1)}}}$ $R_1/R_2=sqrt(M_(r(2))/(M_(r(1)))$

$M_{r} [C l_{2}] = (35.5 \times 2) = 71$ $M_r[Cl_2]=(35.5xx2)=71$

$M_{r} [N_{2}] = (14 \times 2) = 28$ $M_r[N_2]=(14xx2)=28$

So $\frac{R_{C l_{2}}}{R_{N_{2}}}$ $R_(Cl_2)/(R_(N_2)$ $= \sqrt{\frac{28}{71}} = 0.63$ $=sqrt((28)/(71))=0.63$

Answer 2 · 2015-04-18T17:48:27

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

$r_{eff} \propto \frac{1}{\sqrt{M_{M}}}$ $r_"eff" prop 1/sqrt(M_M)$

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

$r_{chlorine} \propto \frac{1}{\sqrt{M_{M chlorine}}}$ $r_"chlorine" prop 1/sqrt(M_("M chlorine"))$

Likewise, the rate of effusion of nitrogen gas will be

$r_{nitrogen} \propto \frac{1}{\sqrt{M_{M nitrogen}}}$ $r_"nitrogen" prop 1/sqrt(M_("M nitrogen"))$

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

$\frac{r_{chlorine}}{r_{nitrogen}} = \frac{\frac{1}{\sqrt{M_{M chlorine}}}}{\frac{1}{\sqrt{M_{M nitrogen}}}} = \frac{\sqrt{M_{M nitrogen}}}{\sqrt{M_{M chlorine}}}$ $r_"chlorine"/r_"nitrogen" = (1/sqrt(M_("M chlorine")))/(1/sqrt(M_("M nitrogen"))) = sqrt(M_("M nitrogen"))/sqrt(M_("M chlorine"))$

The numerical value of this ratio will be

$\frac{r_{chlorine}}{r_{nitrogen}} = \frac{\sqrt{28.014 g/mol}}{\sqrt{70.906 g/mol}} = 0.629$ $r_"chlorine"/r_"nitrogen" = (sqrt(28.014cancel("g/mol")))/(sqrt(70.906cancel("g/mol"))) = color(green)(0.629)$

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

Video from: Noel Pauller

Hope this helps!

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