It takes \(12.6 \mathrm{~s}\) for \(1.73 \times 10^{-3} \mathrm{~mol}\) of \(\mathrm{CO}\) to effuse through a pinhole. Under the same conditions, how long will it take for the same amount of \(\mathrm{CO}_{2}\) to effuse through the same pinhole?

Short Answer

Expert verified
Answer: It will take approximately 10.08 seconds for the same amount of CO2 to effuse through the same pinhole.

Step by step solution

01

Identify necessary information

First, we should gather the needed information to solve the problem. From the exercise, we know the following: - It takes 12.6 seconds for \(1.73 \times 10^{-3} \mathrm{~mol}\) of CO to effuse through a pinhole. - We need to determine the time it takes for the same amount (\(1.73 \times 10^{-3} \mathrm{~mol}\)) of CO2 to effuse through the same pinhole. - We'll use Graham's Law of Effusion to relate the effusion rates of CO and CO2.
02

Utilize Graham's Law of Effusion

Graham's Law of Effusion states that the ratio of the effusion rates of two gases is inversely proportional to the square root of their molar masses. Mathematically, this can be represented as: \(\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}\) where \(r_1\) and \(r_2\) are the effusion rates of gas 1 and gas 2, and \(M_1\) and \(M_2\) are their molar masses. In our case, gas 1 is CO and gas 2 is CO2.
03

Calculate molar masses of CO and CO2

Next, we need to find the molar masses of CO and CO2. For CO: Molar mass of C = 12.01 g/mol Molar mass of O = 16.00 g/mol Molar mass of CO = \((12.01 + 16.00) \mathrm{~g/mol} = 28.01 \mathrm{~g/mol}\) For CO2: Molar mass of C = 12.01 g/mol Molar mass of O = 16.00 g/mol Molar mass of CO2 = \((12.01 + 2(16.00)) \mathrm{~g/mol} = 44.01 \mathrm{~g/mol}\)
04

Apply Graham's Law formula to compute effusion rate ratio

Now we can apply the formula of Graham's Law for effusion rate ratio. By plugging in the molar masses for CO and CO2, we get: \(\frac{r_{CO}}{r_{CO2}} = \sqrt{\frac{M_{CO2}}{M_{CO}}} = \sqrt{\frac{44.01}{28.01}}\) Now, compute the ratio: \(\frac{r_{CO}}{r_{CO2}} \approx 1.25\)
05

Find the time it takes for CO2 to effuse through the pinhole

Recall that the effusion rate is equal to the number of moles divided by the time (r = moles/time). From the exercise, we know it takes 12.6 seconds for \(1.73 \times 10^{-3} \mathrm{~mol}\) of CO to effuse. Since effusion rates are inversely proportional to time, we can rearrange our equation and relate the amount of time for CO and CO2 to effuse as follows: \(\frac{t_{CO}}{t_{CO2}} \approx 1.25\) By plugging in the given time for CO, we can find the time for CO2 effusing through the pinhole: \(t_{CO2} = \frac{t_{CO}}{1.25} \approx \frac{12.6 \mathrm{~s}}{1.25} \approx 10.08 \mathrm{~s}\)
06

Final Answer

Under the same conditions, it will take approximately \(10.08 \mathrm{~s}\) for the same amount of CO2 to effuse through the same pinhole.

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Most popular questions from this chapter

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