Oxygen gas generated by the decomposition of potassium chlorate is collected. The volume of oxygen collected at `24^(@)C` and atmospheric pressure of `760 m m Hg` is `128 mL`. Calculate the mass `(` in grams `)` of oxygen gas obtained. The pressure of water vapour at `24^(@)C` is `22.4 m m Hg`.

Options

A. 1.36 g
B. 1.52 g
C. 0.163 g
D. 1.63 g

Explanation

To solve the problem of calculating the mass of oxygen gas obtained from the decomposition of potassium chlorate, we will follow these steps:

### Step 1: Calculate the pressure of oxygen gas
The total atmospheric pressure is given as 760 mm Hg. The pressure of water vapor at 24°C is 22.4 mm Hg. To find the pressure of the oxygen gas, we subtract the vapor pressure of water from the atmospheric pressure.

\[ P_{\text{O}_2} = P_{\text{total}} - P_{\text{H}_2O} = 760 \, \text{mm Hg} - 22.4 \, \text{mm Hg} = 737.6 \, \text{mm Hg} \]

### Step 2: Convert the pressure of oxygen to atmospheres
To use the ideal gas law, we need the pressure in atmospheres. We convert mm Hg to atm by dividing by 760 mm Hg.

\[ P_{\text{O}_2} = \frac{737.6 \, \text{mm Hg}}{760 \, \text{mm Hg/atm}} = 0.9705 \, \text{atm} \]

### Step 3: Convert the volume of oxygen from mL to L
The volume of oxygen collected is given as 128 mL. We convert this to liters by dividing by 1000.

\[ V = \frac{128 \, \text{mL}}{1000} = 0.128 \, \text{L} \]

### Step 4: Convert the temperature to Kelvin
The temperature is given as 24°C. To convert this to Kelvin, we add 273 to the Celsius temperature.

\[ T = 24 + 273 = 297 \, \text{K} \]

### Step 5: Use the ideal gas law to find the number of moles of oxygen
The ideal gas law is given by the equation \( PV = nRT \). Rearranging this to find the number of moles \( n \):

\[ n = \frac{PV}{RT} \]

Where:
- \( R = 0.0821 \, \text{L atm/(mol K)} \)
- \( P = 0.9705 \, \text{atm} \)
- \( V = 0.128 \, \text{L} \)
- \( T = 297 \, \text{K} \)

Substituting in the values:

\[ n = \frac{(0.9705 \, \text{atm})(0.128 \, \text{L})}{(0.0821 \, \text{L atm/(mol K)})(297 \, \text{K})} \]

Calculating this gives:

\[ n \approx \frac{0.1244}{24.4857} \approx 0.00508 \, \text{mol} \]

### Step 6: Calculate the mass of oxygen
The molecular weight of oxygen (O₂) is 32 g/mol. To find the mass \( W \):

\[ W = n \times M \]

Where \( M \) is the molar mass of oxygen.

\[ W = 0.00508 \, \text{mol} \times 32 \, \text{g/mol} \approx 0.16256 \, \text{g} \]

### Step 7: Round the answer
Rounding to three significant figures, the mass of oxygen gas obtained is approximately:

\[ W \approx 0.163 \, \text{g} \]

### Final Answer
The mass of oxygen gas obtained is **0.163 grams**.

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