The density of Carbon dioxide at 6.0 atm pressure and `27^0C` is :

Options

A. 8.9g`L^(−1)`
B. 1.9g`L^(−1)`
C. 10.7g`L^(−1)`
D. 0.9g`L^(−1)`

Explanation

To find the density of carbon dioxide (CO2) at a pressure of 6.0 atm and a temperature of 27°C, we can use the ideal gas equation and the relationship between density, pressure, and temperature.

### Step-by-Step Solution:

1. **Understand the Ideal Gas Law**: The ideal gas law is given by the equation:
\[ PV = nRT \]
where:
- \( P \) = pressure
- \( V \) = volume
- \( n \) = number of moles
- \( R \) = ideal gas constant
- \( T \) = temperature in Kelvin

2. **Convert Temperature to Kelvin**: The temperature given is 27°C. To convert this to Kelvin:
\[ T(K) = 27 + 273 = 300 \, K \]

3. **Relate Moles to Mass**: The number of moles \( n \) can be expressed in terms of mass \( W \) and molar mass \( M \):
\[ n = \frac{W}{M} \]
Substituting this into the ideal gas equation gives:
\[ PV = \frac{W}{M}RT \]

4. **Rearranging for Density**: Rearranging the equation to express density \( D \) (where density \( D = \frac{W}{V} \)):
\[ D = \frac{PM}{RT} \]

5. **Substituting Known Values**:
- Pressure \( P = 6.0 \, atm \)
- Molar mass of CO2 \( M = 44 \, g/mol \)
- Gas constant \( R = 0.0821 \, \frac{atm \cdot L}{K \cdot mol} \)
- Temperature \( T = 300 \, K \)

Plugging these values into the density equation:
\[ D = \frac{6.0 \, atm \times 44 \, g/mol}{0.0821 \, \frac{atm \cdot L}{K \cdot mol} \times 300 \, K} \]

6. **Calculating the Density**:
\[ D = \frac{264 \, g}{24.63 \, L} \approx 10.7 \, g/L \]

Thus, the density of carbon dioxide at 6.0 atm pressure and 27°C is approximately **10.7 g/L**.