10 " mL of " a gaseous organic compound containing C, H and O only was mixed with 100 " mL of " `O_(2)` and exploded under condition which allowed the `H_(2)O` formed to condense. The volume of the gas after explosion was 90 mL. On treatment with KOH solution, a further contraction of 20 mL in volume was observed. The vapour density of the compound is 23. All volume measurements were made under the same condition.
Q.The volume of `CO_(2)` is

Options

A. 20 mL
B. 50 mL
C. 70 mL
D. 90 mL

Explanation

To solve the problem step by step, let's break down the information given and analyze it carefully.

### Step 1: Understand the Initial Conditions
We have:
- 10 mL of a gaseous organic compound (C, H, O).
- 100 mL of O₂ mixed with the compound.

### Step 2: Analyze the Explosion
After the explosion:
- The total volume of gas is 90 mL.
- This volume consists of the volume of CO₂ produced and any unreacted O₂.

### Step 3: Determine the Contraction After KOH Treatment
When the gas mixture is treated with KOH, a further contraction of 20 mL is observed. KOH absorbs CO₂, so this contraction indicates the volume of CO₂ present in the gas mixture.

### Step 4: Calculate the Volume of CO₂
From the contraction of 20 mL:
- Volume of CO₂ = 20 mL.

### Step 5: Calculate the Volume of Unreacted O₂
Since the total volume after the explosion is 90 mL and we know the volume of CO₂ is 20 mL, we can find the volume of unreacted O₂:
- Volume of unreacted O₂ = Total volume after explosion - Volume of CO₂
- Volume of unreacted O₂ = 90 mL - 20 mL = 70 mL.

### Step 6: Calculate the Volume of O₂ that Reacted
To find the volume of O₂ that reacted with the organic compound:
- Volume of O₂ reacted = Initial volume of O₂ - Volume of unreacted O₂
- Volume of O₂ reacted = 100 mL - 70 mL = 30 mL.

### Final Answer
The volume of CO₂ produced during the reaction is **20 mL**.