90 mL of pure dry `O_(2)` is subjected to silent electric discharge. If only `10%` of it is converted to `O_(3)` volume of the mixture of gases `(O_(2)and O_(3))` after the reaction will be ------ and after passing through alkaline pyrogallol (absorbing agent of oxygen gas) will be

Options

A. 87 mL and 6 mL
B. 81 mL and 87 mL
C. 78 mL and 84 mL
D. 87 mL and 81 mL

Answer

To solve the problem step by step, we will follow the given data and perform the necessary calculations.

### Step 1: Understand the Initial Conditions
We are given:
- Initial volume of oxygen (O₂) = 90 mL
- Percentage of O₂ converted to ozone (O₃) = 10%

### Step 2: Calculate the Volume of O₂ Converted to O₃
To find the volume of O₂ that is converted to O₃, we calculate:
\[ \text{Volume of O₂ converted to O₃} = \frac{10}{100} \times 90 \text{ mL} = 9 \text{ mL} \]

### Step 3: Calculate the Remaining Volume of O₂
Now, we find the volume of O₂ that remains after the conversion:
\[ \text{Remaining volume of O₂} = 90 \text{ mL} - 9 \text{ mL} = 81 \text{ mL} \]

### Step 4: Write the Reaction Equation
The reaction for the conversion of O₂ to O₃ can be represented as:
\[ 3 \text{O₂} \rightarrow 2 \text{O₃} \]
From the equation, we see that 3 volumes of O₂ produce 2 volumes of O₃.

### Step 5: Calculate the Volume of O₃ Produced
Using the stoichiometry of the reaction, we calculate the volume of O₃ produced from the 9 mL of O₂ that was converted:
\[ \text{Volume of O₃} = \frac{2}{3} \times 9 \text{ mL} = 6 \text{ mL} \]

### Step 6: Calculate the Total Volume of the Mixture
Now, we can find the total volume of the gas mixture (O₂ + O₃):
\[ \text{Total volume of mixture} = \text{Remaining O₂} + \text{O₃} = 81 \text{ mL} + 6 \text{ mL} = 87 \text{ mL} \]

### Step 7: Calculate the Volume of O₂ After Passing Through Alkaline Pyrogallol
Alkaline pyrogallol absorbs O₂, so we need to find the volume of O₂ left after passing through it:
\[ \text{Volume of O₂ after passing through alkaline pyrogallol} = \text{Total volume of mixture} - \text{Volume of O₃} = 87 \text{ mL} - 6 \text{ mL} = 81 \text{ mL} \]

### Final Answers
1. Volume of the mixture of gases (O₂ and O₃) after the reaction = **87 mL**
2. Volume of O₂ after passing through alkaline pyrogallol = **81 mL**