Find the mass of `450cm^3` of `N_2` gas at `32^@C` and 750mm of Hg pressure. 

Answer

To find the mass of `450 cm³` of nitrogen gas (`N₂`) at `32°C` and `750 mm Hg` pressure, we can use the ideal gas equation:

\[ PV = nRT \]

Where:
- \( P \) = Pressure
- \( V \) = Volume
- \( n \) = Number of moles
- \( R \) = Universal gas constant
- \( T \) = Temperature in Kelvin

### Step-by-Step Solution:

**Step 1: Convert Volume from cm³ to Liters**
- Given volume = `450 cm³`
- We know that \( 1 cm³ = 10^{-3} L \).

\[ V = 450 \, cm³ \times 10^{-3} \, L/cm³ = 0.45 \, L \]

**Step 2: Convert Temperature from Celsius to Kelvin**
- Given temperature = `32°C`
- To convert Celsius to Kelvin, use the formula:

\[ T(K) = T(°C) + 273 \]

\[ T = 32 + 273 = 305 \, K \]

**Step 3: Convert Pressure from mm Hg to Atmospheres**
- Given pressure = `750 mm Hg`
- We know that \( 1 atm = 760 mm Hg \).

\[ P = \frac{750 \, mm Hg}{760 \, mm Hg/atm} = \frac{750}{760} \, atm \approx 0.9868 \, atm \]

**Step 4: Calculate the Number of Moles (n) using the Ideal Gas Equation**
- Rearranging the ideal gas equation to find \( n \):

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

Substituting the values:
- \( P = 0.9868 \, atm \)
- \( V = 0.45 \, L \)
- \( R = 0.0821 \, L \cdot atm/(mol \cdot K) \)
- \( T = 305 \, K \)

\[ n = \frac{0.9868 \, atm \times 0.45 \, L}{0.0821 \, L \cdot atm/(mol \cdot K) \times 305 \, K} \]

Calculating \( n \):

\[ n \approx \frac{0.4431}{25.1065} \approx 0.0176 \, mol \]

**Step 5: Calculate the Mass of Nitrogen Gas**
- The molecular weight of nitrogen gas \( N_2 \) is \( 28 \, g/mol \) (since the atomic weight of nitrogen is \( 14 \, g/mol \)).

\[ \text{Mass} (M) = n \times \text{Molecular Weight} \]

\[ M = 0.0176 \, mol \times 28 \, g/mol \approx 0.4928 \, g \]

### Final Answer:
The mass of `450 cm³` of nitrogen gas at `32°C` and `750 mm Hg` pressure is approximately **0.493 g**.