At `27^(@)C` and 6.0 atm pressure, the density of propene gas is :

Options

A. 10.2g`L^(−1)`
B. 0.1g`L^(−1)`
C. 1.1g`L^(−1)`
D. 9g`L^(−1)`

Explanation

To find the density of propene gas at 27°C and 6.0 atm pressure, we can use the ideal gas law and the formula for density. Let's go through the steps:

### Step-by-Step Solution:

1. **Understand the Ideal Gas Law**:
The ideal gas law is given by the equation:
\[ PV = nRT \]
where:
- \( P \) = pressure (in atm)
- \( V \) = volume (in liters)
- \( n \) = number of moles
- \( R \) = ideal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = 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. **Calculate Density**:
Density (\( D \)) can be expressed as:
\[ D = \frac{mass}{volume} = \frac{n \cdot M}{V} \]
where \( M \) is the molar mass. From the ideal gas law, we can express \( n \) as:
\[ n = \frac{PV}{RT} \]
Substituting this into the density formula gives:
\[ D = \frac{PM}{RT} \]

4. **Substitute Known Values**:
- \( P = 6.0 \, atm \)
- \( M \) (molar mass of propene, C3H6) = 42 g/mol
- \( R = 0.0821 \, L \cdot atm/(K \cdot mol) \)
- \( T = 300 \, K \)

Plugging these values into the density formula:
\[ D = \frac{(6.0 \, atm)(42 \, g/mol)}{(0.0821 \, L \cdot atm/(K \cdot mol))(300 \, K)} \]

5. **Calculate the Density**:
Performing the calculation:
\[ D = \frac{252 \, g \cdot atm/mol}{24.63 \, L \cdot atm/(K \cdot mol)} \approx 10.2 \, g/L \]

6. **Final Answer**:
The density of propene gas at 27°C and 6.0 atm pressure is approximately:
\[ \text{Density} \approx 10.2 \, g/L \]