Question

A 12.0-liter container holds 709 g of chlorine gas, Cl2, at 35.0 °C. What is the pressure in the container?

21.1 atm

2.39 atm

170 atm

1490 atm

Answer 1

To find the pressure of the chlorine gas in the container, we can use the Ideal Gas Law:

\[ PV = nRT \]

Where:
- \( P \) is the pressure in atm,
- \( V \) is the volume in liters,
- \( n \) is the number of moles of gas,
- \( R \) is the ideal gas constant (0.0821 L·atm/(K·mol)),
- \( T \) is the temperature in Kelvin.

First, we need to convert the given mass of chlorine gas (Cl₂) to moles:

1. Calculate the molar mass of Cl₂:
- The atomic mass of chlorine (Cl) is approximately 35.45 g/mol.
- Therefore, the molar mass of Cl₂ = \( 2 \times 35.45 \Rightarrow 70.90 \) g/mol.

2. Calculate the number of moles (\( n \)):
\[
n = \frac{\text{mass}}{\text{molar mass}} = \frac{709 \text{ g}}{70.90 \text{ g/mol}} \approx 10.01 \text{ moles}
\]

3. Convert the temperature from Celsius to Kelvin:
\[
T(K) = 35.0 + 273.15 = 308.15 \text{ K}
\]

4. Now, we can rearrange the Ideal Gas Law to solve for pressure \( P \):
\[
P = \frac{nRT}{V}
\]

5. Substitute the values into the equation:
- \( n = 10.01 \) moles,
- \( R = 0.0821 \text{ L·atm/(K·mol)} \),
- \( T = 308.15 \text{ K} \),
- \( V = 12.0 \text{ L} \).

\[
P = \frac{(10.01 \text{ mol}) \times (0.0821 \text{ L·atm/(K·mol)}) \times (308.15 \text{ K})}{12.0 \text{ L}}
\]

Calculating \( P \):

\[
P = \frac{(10.01) \times (0.0821) \times (308.15)}{12.0}
\]

Calculating the numerator:

\[
= 10.01 \times 0.0821 \times 308.15 \approx 248.42
\]

Now, divide by 12.0 L:

\[
P \approx \frac{248.42}{12.0} \approx 20.70 \text{ atm}
\]

Therefore, the pressure in the container is approximately **21.1 atm**.

So the correct answer from the options provided is **21.1 atm**.