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 Ideal Gas Law and the relationship between density and molecular weight. Here’s a step-by-step solution:

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

### Step 2: Convert the temperature to Kelvin
The temperature given is \( 100^\circ C \). To convert to Kelvin:
\[ T(K) = T(°C) + 273.15 \]
\[ T = 100 + 273.15 = 373.15 \, K \]

### Step 3: Relate number of moles to mass and molecular weight
The number of moles \( n \) can be expressed in terms of mass \( w \) and molecular weight \( M \):
\[ n = \frac{w}{M} \]
Substituting this into the Ideal Gas Law gives:
\[ PV = \frac{w}{M} RT \]

### Step 4: Rearrange the equation to find molecular weight
Rearranging the equation to isolate \( M \):
\[ M = \frac{wRT}{PV} \]

### Step 5: Use the density to express mass
Density (\( d \)) is defined as mass per unit volume:
\[ d = \frac{w}{V} \]
Thus, mass can be expressed as:
\[ w = dV \]
Substituting this into the molecular weight equation gives:
\[ M = \frac{dVRT}{PV} \]
The volume \( V \) cancels out:
\[ M = \frac{dRT}{P} \]

### Step 6: Substitute the known values
Now we can substitute the known values into the equation:
- Density \( d = 1.15 \, g/L \)
- \( R = 0.0821 \, L·atm/(K·mol) \)
- \( T = 373.15 \, K \)
- \( P = 0.80 \, atm \)

Substituting these values:
\[ M = \frac{(1.15 \, g/L)(0.0821 \, L·atm/(K·mol))(373.15 \, K)}{0.80 \, atm} \]

### Step 7: Calculate the molecular weight
Calculating the right-hand side:
\[ M = \frac{(1.15)(0.0821)(373.15)}{0.80} \]
\[ M = \frac{(28.36)}{0.80} \]
\[ M = 35.45 \, g/mol \]

### Step 8: Final result
The molecular weight of the gas is approximately \( 44 \, g/mol \).