What is the molecular weight of a gas whose density `40^(@)C and 785 mm` of `Hg` pressure is `1.3 g L^(-1)`?

Options

A. `32`
B. `40`
C. `15`
D. `98`

Explanation

To find the molecular weight of the gas, we can use the ideal gas law rearranged in terms of density. The ideal gas law is given by:

\[ PV = nRT \]

Where:
- \( P \) = pressure
- \( V \) = volume
- \( n \) = number of moles
- \( R \) = ideal gas constant
- \( T \) = temperature in Kelvin

We can express the number of moles \( n \) as:

\[ n = \frac{m}{M} \]

Where:
- \( m \) = mass of the gas
- \( M \) = molar mass (molecular weight) of the gas

Substituting this into the ideal gas law gives:

\[ PV = \frac{m}{M}RT \]

Rearranging for \( M \) gives:

\[ M = \frac{mRT}{PV} \]

We can express density \( d \) as:

\[ d = \frac{m}{V} \]

Thus, we can rewrite \( m \) as:

\[ m = dV \]

Substituting this back into the equation for \( M \):

\[ M = \frac{dVRT}{PV} \]

The \( V \) cancels out:

\[ M = \frac{dRT}{P} \]

Now we can substitute the values:

1. **Convert Temperature to Kelvin**:
\[ T = 40^\circ C + 273.15 = 313.15 \, K \]

2. **Convert Pressure from mmHg to atm**:
\[ P = \frac{785 \, mmHg}{760 \, mmHg/atm} = 1.03289 \, atm \]

3. **Use the Ideal Gas Constant**:
\[ R = 0.0821 \, L \cdot atm \cdot K^{-1} \cdot mol^{-1} \]

4. **Substitute the values into the equation**:
\[ M = \frac{(1.3 \, g/L)(0.0821 \, L \cdot atm \cdot K^{-1} \cdot mol^{-1})(313.15 \, K)}{1.03289 \, atm} \]

5. **Calculate**:
\[ M = \frac{(1.3)(0.0821)(313.15)}{1.03289} \]
\[ M = \frac{33.198}{1.03289} \]
\[ M \approx 32.1 \, g/mol \]

Thus, the molecular weight of the gas is approximately **32 g/mol**.