At 1 atm and 273 K the density of gas, whose molecular weight is 45, is:

Options

A. 44.8 g/L
B. 11.4 g/L
C. 2 g/L
D. 3 g/L

Explanation

To find the density of a gas at a given pressure and temperature, we can use the formula:

\[ \text{Density} (d) = \frac{P \times M}{R \times T} \]

Where:
- \( P \) = pressure in atm
- \( M \) = molar mass (molecular weight) in grams per mole
- \( R \) = ideal gas constant = 0.0821 L·atm/(K·mol)
- \( T \) = temperature in Kelvin

### Step-by-Step Solution:

**Step 1: Identify the given values.**
- Pressure \( P = 1 \) atm
- Molecular weight \( M = 45 \) g/mol
- Temperature \( T = 273 \) K
- Ideal gas constant \( R = 0.0821 \) L·atm/(K·mol)

**Step 2: Substitute the values into the density formula.**
\[ d = \frac{1 \, \text{atm} \times 45 \, \text{g/mol}}{0.0821 \, \text{L·atm/(K·mol)} \times 273 \, \text{K}} \]

**Step 3: Calculate the denominator.**
\[ 0.0821 \times 273 \approx 22.4133 \]

**Step 4: Substitute the calculated denominator back into the formula.**
\[ d = \frac{45}{22.4133} \]

**Step 5: Perform the division.**
\[ d \approx 2.0067 \, \text{g/L} \]

**Step 6: Round off the answer.**
The density of the gas is approximately \( 2 \, \text{g/L} \).

### Final Answer:
The density of the gas at 1 atm and 273 K is approximately \( 2 \, \text{g/L} \).

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