Answer: The amount of time required by iodine gas to effuse is 112.9 seconds.
Explanation:
Rate of a gas is defined as the amount of gas displaced in a given amount of time.
[tex]\text{Rate}=\frac{n}{t}[/tex]
To calculate the rate of diffusion of gas, we use Graham's Law.
This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows the equation:
[tex]\text{Rate of effusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}[/tex]
So,
[tex]\left(\frac{\frac{n_{N_2O}}{t_{N_2O}}}{\frac{n_{I_2}}{t_{I_2}}}\right)=\sqrt{\frac{M_{I_2}}{M_{N_2O}}}[/tex]
We are given:
Moles of iodine gas = Moles of nitrous oxide gas
Time taken by nitrous oxide gas = 47 seconds
Molar mass of iodine gas = 254 g/mol
Molar mass of nitrous oxide gas = 44 g/mol
Putting values in above equation, we get:
[tex]\left(\frac{\frac{n_{I_2}}{47}}{\frac{n_{I_2}}{t_{I_2}}}\right)=\sqrt{\frac{254}{44}}\\\\\frac{t_{I_2}}{47}=2.403\\\\t=112.9s[/tex]
Hence, the amount of time required by iodine gas to effuse is 112.9 seconds.
it will take 112.9 s for iodine gas, I₂ to effuse from the container under identical conditions
From the question given above, the following data were obtained:
Time for N₂O (t₁) = 47 s
Molar mass of N₂O (M₁) = (2×14) + 16 = 44 g/mol
Molar mass of I₂ (M₂) = 2 × 127 = 254 g/mol
Time for I₂ (t₂) =?
Using the Graham's law of diffusion, we can obtain the time take for iodine gas, I₂ to effuse from the container. This is illustrated below:
[tex]\frac{t_{2}}{t_{1 }} = \sqrt{\frac{M_{2}}{M_{1 }} }\\\\\frac{t_{2}}{47} = \sqrt{\frac{254}{44} }\\\\[/tex]
Cross multiply
[tex]t_{2} = 47 \sqrt{\frac{254}{44} }\\\\[/tex]
Therefore, it will take 112.9 s for iodine gas, I₂ to effuse from the container.
Learn more: https://brainly.com/question/14545265
