Question

If it takes 10.6 hours for 1.00 L of nitrogen, N2, to effuse through the pores in a balloon, how long would it take for 1.00 L of chlorine, Cl2, to effuse under the same conditions?

Answer 1

Answer : The time taken for 1.00 L of chlorine to effuse under the same conditions is, 16.9 hr

Explanation :

According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.

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

And the relation between the rate of effusion and volume is :

$R=\frac{V}{t}$

From this we conclude that,  at constant volume and rate the relation between time and molar mass of gas will be:

$\frac{t_{Cl_2}}{t_{N_2}}=\sqrt{\frac{M_{Cl_2}}{M_{N_2}}}$

where,

$t_{Cl_2}$ = time of $Cl_2$ = ?

$t_{N_2}$ = time of $N_2$ = 10.6 hr

$M_{Cl_2}$ = molar mass of $Cl_2$ = 71 g/mol

$M_{N_2}$ = molar mass of $N_2$ = 28 g/mol

Now put all the given values in this expression, we get:

$\frac{t_{Cl_2}}{10.6hr}=\sqrt{\frac{71g/mol}{28g/mol}}$

$t_{Cl_2}=16.9hr$

Therefore, the time taken for 1.00 L of chlorine to effuse under the same conditions is, 16.9 hr