Question
If 1.01 mol of CO occupies 14.8 L, how many moles of CO will occupy a 73.9 L container at the same temperature and pressure? Round your answer to 3 significant figures.
Answers
GPT-4o mini
To find how many moles of CO will occupy a 73.9 L container at the same temperature and pressure, we can use the concept of molar volume.
From the given data, we know that:
- 1.01 mol of CO occupies 14.8 L.
We can first calculate the molar volume of CO:
\[
\text{Molar Volume} = \frac{\text{Number of moles}}{\text{Volume}} = \frac{1.01 \, \text{mol}}{14.8 \, \text{L}} \approx 0.068 ( \text{mol/L} )
\]
Now, we can use this molar volume to determine how many moles will occupy 73.9 L:
\[
\text{Number of moles} = \text{Volume} \times \text{Molar concentration}
\]
\[
\text{Number of moles} = \frac{73.9 \, \text{L}}{14.8 \, \text{L}} \times 1.01 \, \text{mol} \approx 5.01 \, \text{mol}
\]
So, we need to know how many moles occupy 73.9 L, and that can also be calculated in a straightforward ratio since the temperature and pressure are constant.
Using the ratio of volumes:
\[
\text{Moles in 73.9 L} = 1.01 \, \text{mol} \times \frac{73.9 \, \text{L}}{14.8 \, \text{L}}
\]
Calculating this gives:
\[
\text{Moles in 73.9 L} \approx 1.01 \times 4.99 \approx 5.02 \, \text{mol}
\]
Rounding to three significant figures, the answer is:
\[
\boxed{5.02} \, \text{mol}
\]
From the given data, we know that:
- 1.01 mol of CO occupies 14.8 L.
We can first calculate the molar volume of CO:
\[
\text{Molar Volume} = \frac{\text{Number of moles}}{\text{Volume}} = \frac{1.01 \, \text{mol}}{14.8 \, \text{L}} \approx 0.068 ( \text{mol/L} )
\]
Now, we can use this molar volume to determine how many moles will occupy 73.9 L:
\[
\text{Number of moles} = \text{Volume} \times \text{Molar concentration}
\]
\[
\text{Number of moles} = \frac{73.9 \, \text{L}}{14.8 \, \text{L}} \times 1.01 \, \text{mol} \approx 5.01 \, \text{mol}
\]
So, we need to know how many moles occupy 73.9 L, and that can also be calculated in a straightforward ratio since the temperature and pressure are constant.
Using the ratio of volumes:
\[
\text{Moles in 73.9 L} = 1.01 \, \text{mol} \times \frac{73.9 \, \text{L}}{14.8 \, \text{L}}
\]
Calculating this gives:
\[
\text{Moles in 73.9 L} \approx 1.01 \times 4.99 \approx 5.02 \, \text{mol}
\]
Rounding to three significant figures, the answer is:
\[
\boxed{5.02} \, \text{mol}
\]