A sample of a gas at `100^(@)C` and 0.80 atm pressure has a density of 1.15 g `L^(-1)`. What is the molecular weight of the gas?

Options

A. `88.0`
B. `44.0`
C. `28.0`
D. `46.0`

Explanation

To find the molecular weight of the gas, we can use the relationship between density, pressure, temperature, and molecular weight derived from the ideal gas law. Here’s a step-by-step solution:

### Step 1: Write down the ideal gas equation
The ideal gas equation is given by:
\[ PV = nRT \]
Where:
- \( P \) = pressure (in atm)
- \( V \) = volume (in liters)
- \( n \) = number of moles of gas
- \( R \) = ideal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = temperature (in Kelvin)

### Step 2: Relate moles to mass and molecular weight
The number of moles \( n \) can also be expressed in terms of mass \( m \) and molecular weight \( M \):
\[ n = \frac{m}{M} \]
Substituting this into the ideal gas equation gives:
\[ PV = \frac{m}{M} RT \]

### Step 3: Rearrange the equation to find molecular weight
Rearranging the equation to solve for molecular weight \( M \):
\[ M = \frac{mRT}{PV} \]

### Step 4: Express mass in terms of density
Using the definition of density \( d \) (density = mass/volume), we can express mass as:
\[ m = dV \]
Substituting this into the equation for molecular weight:
\[ M = \frac{dVRT}{PV} \]
The volume \( V \) cancels out:
\[ M = \frac{dRT}{P} \]

### Step 5: Substitute the known values
Now, we can substitute the known values:
- Density \( d = 1.15 \, \text{g/L} \)
- Temperature \( T = 100^\circ C = 373 \, \text{K} \) (convert Celsius to Kelvin)
- Pressure \( P = 0.80 \, \text{atm} \)
- Gas constant \( R = 0.0821 \, \text{L·atm/(K·mol)} \)

Substituting these values into the equation:
\[ M = \frac{(1.15 \, \text{g/L}) \times (0.0821 \, \text{L·atm/(K·mol)}) \times (373 \, \text{K})}{0.80 \, \text{atm}} \]

### Step 6: Calculate the molecular weight
Now, performing the calculation:
1. Calculate \( 1.15 \times 0.0821 \times 373 \):
\[ 1.15 \times 0.0821 \times 373 = 36.11 \, \text{g·atm/(K·mol)} \]
2. Divide by \( 0.80 \):
\[ M = \frac{36.11}{0.80} = 45.14 \, \text{g/mol} \]

### Step 7: Round to appropriate significant figures
The molecular weight is approximately:
\[ M \approx 44 \, \text{g/mol} \]

### Final Answer
The molecular weight of the gas is \( 44 \, \text{g/mol} \).

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