`1470 cm^3` of a gas is collected over water at 303 K and 74.4 cm of Hg. If the gas weighs 1.98 g and vapour pressure of water at `30^@C` is 3.2 cm of Hg, calculate the molecular weight of the gas.

Explanation

To calculate the molecular weight of the gas collected over water, we can follow these steps:

### Step 1: Identify the Given Data
- Volume of gas (V) = 1470 cm³
- Temperature (T) = 303 K
- Total pressure (P) = 74.4 cm of Hg
- Vapor pressure of water (p) at 30°C = 3.2 cm of Hg
- Weight of gas (W) = 1.98 g

### Step 2: Convert Volume to Liters
To convert the volume from cm³ to liters:
\[ V = \frac{1470 \, \text{cm}^3}{1000} = 1.47 \, \text{L} \]

### Step 3: Calculate the Pressure of the Gas
The pressure of the gas (P_g) can be calculated by subtracting the vapor pressure of water from the total pressure:
\[ P_g = P - p = 74.4 \, \text{cm of Hg} - 3.2 \, \text{cm of Hg} = 71.2 \, \text{cm of Hg} \]

### Step 4: Convert Pressure from cm of Hg to Pascal
To convert the pressure from cm of Hg to Pascal, we use the conversion factor (1 cm of Hg = 133.322 Pa):
\[ P_g = 71.2 \, \text{cm of Hg} \times \frac{133.322 \, \text{Pa}}{1 \, \text{cm of Hg}} = 9486.38 \, \text{Pa} \]

### Step 5: Use the Ideal Gas Law
The ideal gas law is given by:
\[ PV = nRT \]
Where:
- \( P \) = pressure of the gas in Pascal
- \( V \) = volume of the gas in liters
- \( n \) = number of moles of gas
- \( R \) = universal gas constant = 0.0821 L·atm/(K·mol) or 8.314 J/(K·mol)
- \( T \) = temperature in Kelvin

We can express the number of moles (n) in terms of weight (W) and molecular weight (M):
\[ n = \frac{W}{M} \]

Substituting into the ideal gas equation:
\[ P_g \cdot V = \frac{W}{M} \cdot R \cdot T \]

### Step 6: Rearranging for Molecular Weight (M)
Rearranging the equation to solve for M:
\[ M = \frac{W \cdot R \cdot T}{P_g \cdot V} \]

### Step 7: Substitute Values
Substituting the known values:
- \( W = 1.98 \, \text{g} \)
- \( R = 0.0821 \, \text{L·atm/(K·mol)} \) (we need to convert it to match the units of pressure, so we will use \( R = 8.314 \, \text{J/(K·mol)} \))
- \( T = 303 \, \text{K} \)
- \( P_g = 9486.38 \, \text{Pa} \)
- \( V = 1.47 \, \text{L} = 1.47 \times 10^{-3} \, \text{m}^3 \)

Now substituting:
\[ M = \frac{1.98 \, \text{g} \cdot 8.314 \, \text{J/(K·mol)} \cdot 303 \, \text{K}}{9486.38 \, \text{Pa} \cdot 1.47 \times 10^{-3} \, \text{m}^3} \]

### Step 8: Calculate M
Calculating the above expression:
\[ M \approx \frac{1.98 \cdot 8.314 \cdot 303}{9486.38 \cdot 1.47 \times 10^{-3}} \]
\[ M \approx 35.76 \, \text{g/mol} \]

### Conclusion
The molecular weight of the gas is approximately **35.76 g/mol**.

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