What is the density of chlorine gas at 1.21 atm and 34.9 degree Celsius?

To find the density of chlorine gas (\(Cl_2\)) at a given pressure and temperature, we can use the Ideal Gas Law, which is given by the equation:

\[
PV = nRT
\]

Where:
- \(P\) = pressure (in atm)
- \(V\) = volume (in liters)
- \(n\) = number of moles of gas
- \(R\) = ideal gas constant \((0.0821 \, \text{L} \cdot \text{atm} / \text{K} \cdot \text{mol})\)
- \(T\) = temperature (in Kelvin)

### Step 1: Convert the temperature to Kelvin

To convert the temperature from Celsius to Kelvin, use the formula:

\[
T(K) = T(°C) + 273.15
\]

\[
T = 34.9 + 273.15 = 308.05 \, \text{K}
\]

### Step 2: Use the Ideal Gas Law

Rearranging the Ideal Gas Law to find the density \((\rho)\):

\[
n = \frac{PV}{RT}
\]

We also know that the number of moles (\(n\)) can be expressed in terms of density as:

\[
n = \frac{m}{M}
\]

Where:
- \(m\) = mass of the gas (in grams)
- \(M\) = molar mass of the gas (in g/mol)

Substituting this into the equation gives:

\[
\frac{m}{M} = \frac{PV}{RT}
\]

### Step 3: Rearranging to find density

From this, we can express density (\(\rho\)) as:

\[
\rho = \frac{m}{V} = \frac{P \cdot M}{R \cdot T}
\]

### Step 4: Find the molar mass of chlorine gas

The molar mass of chlorine gas (\(Cl_2\)) is:

\[
M = 2 \times 35.453 \, \text{g/mol} = 70.906 \, \text{g/mol}
\]

### Step 5: Plug in the values

Now, we can plug in the values into the density formula:

- \(P = 1.21 \, \text{atm}\)
- \(M = 70.906 \, \text{g/mol}\)
- \(R = 0.0821 \, \text{L} \cdot \text{atm} / \text{K} \cdot \text{mol}\)
- \(T = 308.05 \, \text{K}\)

\[
\rho = \frac{1.21 \, \text{atm} \times 70.906 \, \text{g/mol}}{0.0821 \, \text{L} \cdot \text{atm} / \text{K} \cdot \text{mol} \times 308.05 \, \text{K}}
\]

### Step 6: Calculate the density

Now, let’s perform the calculation:

\[
\rho = \frac{1.21 \times 70.906}{0.0821 \times 308.05}
\]

Calculating the values:

1. Calculate the numerator: \(1.21 \times 70.906 \approx 85.89866\)
2. Calculate the denominator: \(0.0821 \times 308.05 \approx 25.358755\)
3. Finally, divide the two results:

\[
\rho \approx \frac{85.89866}{25.358755} \approx 3.39 \, \text{g/L}
\]

### Final Answer

The density of chlorine gas at \(1.21 \, \text{atm}\) and \(34.9^\circ \text{C}\) is approximately **3.39 g/L**.