Calculate the density of `NH_(3)` at `30^(@)C` and 5 atm pressure.

Options

A. `3.42 g L^(-1)`
B. `2.42 g L^(-1)`
C. `1.71 g L^(-1)`
D. `3.84 g L^(-1)`

Explanation

To calculate the density of ammonia (NH₃) at 30°C and 5 atm pressure, we can use the ideal gas equation, which is given by:

\[ PV = nRT \]

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)

### Step-by-Step Solution:

**Step 1: Convert the temperature from Celsius to Kelvin.**
- The formula to convert Celsius to Kelvin is:
\[ T(K) = T(°C) + 273.15 \]
- For 30°C:
\[ T = 30 + 273.15 = 303.15 \, K \]

**Step 2: Identify the values needed for the calculation.**
- Pressure \( P = 5 \, atm \)
- Molar mass of ammonia \( NH₃ \):
- Nitrogen (N) = 14 g/mol
- Hydrogen (H) = 1 g/mol, and since there are 3 hydrogen atoms:
\[ \text{Molar mass of } NH₃ = 14 + (3 \times 1) = 17 \, g/mol \]

**Step 3: Rearrange the ideal gas equation to find density.**
- We know that density \( D \) can be expressed as:
\[ D = \frac{m}{V} \]
- The number of moles \( n \) can be expressed as:
\[ n = \frac{m}{M} \]
where \( M \) is the molar mass.
- Substituting \( n \) into the ideal gas equation:
\[ PV = \frac{m}{M}RT \]
- Rearranging for density \( D \):
\[ D = \frac{PM}{RT} \]

**Step 4: Substitute the known values into the density equation.**
- Substitute \( P = 5 \, atm \), \( M = 17 \, g/mol \), \( R = 0.0821 \, L·atm/(K·mol) \), and \( T = 303.15 \, K \):
\[ D = \frac{5 \, atm \times 17 \, g/mol}{0.0821 \, L·atm/(K·mol) \times 303.15 \, K} \]

**Step 5: Calculate the density.**
- First, calculate the denominator:
\[ 0.0821 \times 303.15 \approx 24.745 \]
- Now calculate \( D \):
\[ D = \frac{85}{24.745} \approx 3.43 \, g/L \]

### Final Answer:
The density of ammonia (NH₃) at 30°C and 5 atm pressure is approximately **3.43 g/L**.