Which of the following gases will have density of `1.8 g L^(-1)` at 760 torr pressure and `27^(@)C`?

Options

A. `O_(2)`
B. `CO_(2)`
C. `NH_(3)`
D. `SO_(2)`

Explanation

To determine which gas has a density of `1.8 g L^(-1)` at a pressure of `760 torr` and a temperature of `27°C`, we will use the ideal gas law and the relationship between density and molar mass.

### Step-by-Step Solution:

1. **Convert Temperature to Kelvin:**
\[ T(K) = T(°C) + 273 = 27 + 273 = 300 \, K \]

**Hint:** Always convert Celsius to Kelvin by adding 273.

2. **Convert Pressure to Atmospheres:**
\[ P(atm) = \frac{760 \, torr}{760} = 1 \, atm \]

**Hint:** Remember that 760 torr is equivalent to 1 atm.

3. **Use the Ideal Gas Law:**
The ideal gas law is given by:
\[ PV = nRT \]
Rearranging for density (\( \rho \)):
\[ \rho = \frac{m}{V} = \frac{n \cdot M}{V} \]
where \( n = \frac{PV}{RT} \) and \( M \) is the molar mass.

4. **Substituting for Density:**
\[ \rho = \frac{PM}{RT} \]

5. **Plugging in Known Values:**
- \( P = 1 \, atm \)
- \( R = 0.0821 \, L \cdot atm \cdot K^{-1} \cdot mol^{-1} \)
- \( T = 300 \, K \)
- \( \rho = 1.8 \, g/L \)

Rearranging for molar mass \( M \):
\[ M = \frac{\rho RT}{P} \]

6. **Calculating Molar Mass:**
\[ M = \frac{1.8 \, g/L \cdot 0.0821 \, L \cdot atm \cdot K^{-1} \cdot mol^{-1} \cdot 300 \, K}{1 \, atm} \]
\[ M = \frac{1.8 \cdot 0.0821 \cdot 300}{1} \]
\[ M = 44.0 \, g/mol \]

7. **Identifying the Gas:**
Now, we check the options for a gas with a molar mass of approximately 44 g/mol. The gas that fits this description is:
- **Option B: \( CO_2 \)** (Carbon Dioxide), which has a molar mass of \( 12 + 16 \times 2 = 44 \, g/mol \).

### Conclusion:
The gas that has a density of `1.8 g L^(-1)` at `760 torr` and `27°C` is **Carbon Dioxide (CO₂)**.

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