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.
[tex]R\propto \sqrt{\frac{1}{M}}[/tex]
And the relation between the rate of effusion and volume is :
[tex]R=\frac{V}{t}[/tex]
From this we conclude that, at constant volume and rate the relation between time and molar mass of gas will be:
[tex]\frac{t_{Cl_2}}{t_{N_2}}=\sqrt{\frac{M_{Cl_2}}{M_{N_2}}}[/tex]
where,
[tex]t_{Cl_2}[/tex] = time of [tex]Cl_2[/tex] = ?
[tex]t_{N_2}[/tex] = time of [tex]N_2[/tex] = 10.6 hr
[tex]M_{Cl_2}[/tex] = molar mass of [tex]Cl_2[/tex] = 71 g/mol
[tex]M_{N_2}[/tex] = molar mass of [tex]N_2[/tex] = 28 g/mol
Now put all the given values in this expression, we get:
[tex]\frac{t_{Cl_2}}{10.6hr}=\sqrt{\frac{71g/mol}{28g/mol}}[/tex]
[tex]t_{Cl_2}=16.9hr[/tex]
Therefore, the time taken for 1.00 L of chlorine to effuse under the same conditions is, 16.9 hr
It will take 16.9 hours for chlorine gas (Cl₂) to effuse under the same condition.
From the question given above, the following data were obtained:
Time for N₂ (t₁) = 10.6 hours
Molar mass of N₂ (M₁) = 2 × 14 = 28 g/mol
Molar mass of Cl₂ (M₂) = 2 × 35.5 = 71 g/mol
Time for Cl₂ (t₂) =?
Using the Graham's law of diffusion, we can obtain the time take for chlorine gas, Cl₂ to effuse as shown below:
[tex]\frac{t_{2}}{t_{1}} = \sqrt{\frac{M_{2}}{M_{1}}} \\\\\frac{t_{2}}{10.6} = \sqrt{\frac{71}{28}}\\\\[/tex]
Cross multiply
[tex]t_{2} = 10.6\sqrt{\frac{71}{28}} \\\\[/tex]
Therefore, it will take 16.9 hours for Cl₂ to effuse under the same condition.
Learn more: https://brainly.com/question/8804761
