A vessel contains a mixtrue consisting of `m_(1) = 7g` of nitrogen `M_(1) = 28 and m_(2) = 11 g` of carbon dioxide `(M_(2) 44)` at temperature` T = 300 K` and pressure `p_(0) = 1` atm. Find the density of the mixture.

Options

A. 1.446 g per litre
B. 2.567 g per litre
C. 3.752 g per litre
D. 4.572 g per litre

Answer

To find the density of the mixture of nitrogen and carbon dioxide, we can use the ideal gas law and the concept of partial pressures. Here’s a step-by-step solution:

### Step 1: Identify the given data
- Mass of nitrogen, \( m_1 = 7 \, \text{g} \)
- Molar mass of nitrogen, \( M_1 = 28 \, \text{g/mol} \)
- Mass of carbon dioxide, \( m_2 = 11 \, \text{g} \)
- Molar mass of carbon dioxide, \( M_2 = 44 \, \text{g/mol} \)
- Temperature, \( T = 300 \, \text{K} \)
- Pressure, \( p_0 = 1 \, \text{atm} \)

### Step 2: Convert the masses to moles
Using the formula:
\[ n = \frac{m}{M} \]
Calculate the number of moles for each gas:
- For nitrogen:
\[ n_1 = \frac{m_1}{M_1} = \frac{7 \, \text{g}}{28 \, \text{g/mol}} = 0.25 \, \text{mol} \]
- For carbon dioxide:
\[ n_2 = \frac{m_2}{M_2} = \frac{11 \, \text{g}}{44 \, \text{g/mol}} = 0.25 \, \text{mol} \]

### Step 3: Calculate the total number of moles
\[ n_{total} = n_1 + n_2 = 0.25 \, \text{mol} + 0.25 \, \text{mol} = 0.5 \, \text{mol} \]

### Step 4: Use the ideal gas law to find the volume
The ideal gas law is given by:
\[ PV = nRT \]
Rearranging for volume \( V \):
\[ V = \frac{nRT}{P} \]
Substituting the values:
- \( R = 0.0821 \, \text{L atm/(K mol)} \)
- Convert pressure to atm (already in atm)
\[ V = \frac{(0.5 \, \text{mol})(0.0821 \, \text{L atm/(K mol)})(300 \, \text{K})}{1 \, \text{atm}} = \frac{12.315}{1} = 12.315 \, \text{L} \]

### Step 5: Calculate the total mass of the mixture
\[ m_{total} = m_1 + m_2 = 7 \, \text{g} + 11 \, \text{g} = 18 \, \text{g} \]

### Step 6: Calculate the density of the mixture
Density \( \rho \) is given by:
\[ \rho = \frac{m_{total}}{V} \]
Substituting the values:
\[ \rho = \frac{18 \, \text{g}}{12.315 \, \text{L}} = 1.46 \, \text{g/L} \]
To convert to kg/m³:
\[ \rho = 1.46 \, \text{g/L} \times 1000 = 1460 \, \text{kg/m}^3 \]

### Final Answer
The density of the mixture is approximately \( 1460 \, \text{kg/m}^3 \).

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