Question

a 10.5 l sample of gas at 292 k contains o₂ at 0.622 atm and n₂ at 0.517 atm. calculate the number of moles of o₂ present. 0.500 mol 0.273 mol 1.00 mol 1.39 mol 2.66 mol

Explanation:

Step1: Recall the Ideal Gas Law

The Ideal Gas Law is given by the formula $PV = nRT$, where $P$ is the pressure, $V$ is the volume, $n$ is the number of moles, $R$ is the ideal gas constant, and $T$ is the temperature in Kelvin. For this problem, we want to solve for $n$ (moles of $\ce{O2}$), so we can rearrange the formula to $n=\frac{PV}{RT}$.

Step2: Identify the values for $\ce{O2}$

• Pressure of $\ce{O2}$ ($P$) = 0.622 atm
• Volume of the gas sample ($V$) = 10.5 L (since the $\ce{O2}$ is part of this sample, it occupies the same volume)
• Temperature ($T$) = 292 K
• Ideal gas constant ($R$) = 0.0821 $\frac{\text{L·atm}}{\text{mol·K}}$ (this is the appropriate value for these units)

Step3: Plug the values into the formula

Substitute the values into the rearranged Ideal Gas Law formula:
$$n=\frac{(0.622\ \text{atm})(10.5\ \text{L})}{(0.0821\ \frac{\text{L·atm}}{\text{mol·K}})(292\ \text{K})}$$

Step4: Calculate the numerator and the denominator

• Numerator: $(0.622\ \text{atm})(10.5\ \text{L}) = 0.622\times10.5 = 6.531\ \text{L·atm}$
• Denominator: $(0.0821\ \frac{\text{L·atm}}{\text{mol·K}})(292\ \text{K}) = 0.0821\times292 \approx 23.9732\ \frac{\text{L·atm}}{\text{mol}}$

Step5: Divide to find $n$

Now, divide the numerator by the denominator:
$$n=\frac{6.531}{23.9732}\approx0.273\ \text{mol}$$

Answer:

0.273 mol