When silent electric discharge is passed through `O_(2)` gas, it converts into `O_(3)` .If the density of the final sample is 20 times the density of hydrogen gas under similar conditions, calculate the mass percent of `O_(2)` in the final sample.

Answer

To solve the problem step by step, we will follow the information provided in the question and the video transcript.

### Step 1: Understanding the Reaction
When silent electric discharge is passed through oxygen gas (O₂), it converts into ozone (O₃). Therefore, we have a mixture of O₂ and O₃ in equilibrium.

### Step 2: Density Information
The density of the final sample is given as 20 times the density of hydrogen gas (H₂). The density of hydrogen gas (H₂) under standard conditions is approximately 0.0899 g/L, but for calculations, we can use the molar mass of hydrogen, which is 2 g/mol.

Thus, the density of the final sample (D) can be calculated as:
\[ D = 20 \times \text{density of } H_2 = 20 \times 0.0899 \, \text{g/L} = 1.798 \, \text{g/L} \]

### Step 3: Calculate the Molar Mass of the Mixture
The density of a gas can also be expressed in terms of its molar mass (M) and the universal gas constant (R) at standard temperature and pressure (STP):
\[ D = \frac{M}{RT} \]
At STP, R = 0.0821 L·atm/(K·mol) and T = 273.15 K. However, we can simplify our calculation by using the relationship:
\[ D = \frac{M}{22.4} \]
where 22.4 L is the molar volume of an ideal gas at STP.

From the density of the mixture:
\[ M = D \times 22.4 = 20 \times 2 = 40 \, \text{g/mol} \]

### Step 4: Molar Mass of O₂ and O₃
The molar mass of O₂ is:
\[ M_{O_2} = 32 \, \text{g/mol} \]
The molar mass of O₃ is:
\[ M_{O_3} = 48 \, \text{g/mol} \]

### Step 5: Setting Up the Equation
Let \( x \) be the number of moles of O₂ and \( y \) be the number of moles of O₃ in the mixture. The average molar mass of the mixture can be expressed as:
\[ \text{Average Molar Mass} = \frac{32x + 48y}{x + y} \]
Setting this equal to the average molar mass we calculated:
\[ \frac{32x + 48y}{x + y} = 40 \]

### Step 6: Solving for the Ratio of Moles
Cross-multiplying gives:
\[ 32x + 48y = 40(x + y) \]
Expanding and rearranging:
\[ 32x + 48y = 40x + 40y \]
\[ 8y = 8x \]
Thus, we find:
\[ y = x \]
This means the number of moles of O₂ is equal to the number of moles of O₃.

### Step 7: Mass Percent Calculation
The total mass of the mixture can be expressed as:
\[ \text{Total mass} = 32x + 48y = 32x + 48x = 80x \]
The mass percent of O₂ in the mixture is then:
\[ \text{Mass percent of } O_2 = \frac{\text{mass of } O_2}{\text{total mass}} \times 100 = \frac{32x}{80x} \times 100 = \frac{32}{80} \times 100 = 40\% \]

### Final Answer
The mass percent of O₂ in the final sample is **40%**.

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