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 unreacted `O_(2)` is

Options

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

Explanation

To solve the problem step by step, we will analyze the information given and apply stoichiometric principles.

### Step 1: Understand the initial conditions
We have:
- Volume of the gaseous organic compound (C, H, O) = 10 mL
- Volume of O₂ mixed = 100 mL
- Total volume after explosion = 90 mL

### Step 2: Determine the volume of gases after the reaction
After the explosion, the volume of gas remaining is 90 mL. This volume consists of unreacted O₂ and the products of the reaction, which include CO₂ and possibly some unreacted organic compound.

### Step 3: Identify the volume of water formed
Since the problem states that the water formed condensed, we can assume that it does not contribute to the gas volume after the explosion.

### Step 4: Calculate the volume of CO₂ formed
The problem mentions that after treatment with KOH, there is a further contraction of 20 mL. KOH absorbs CO₂, so this contraction indicates the volume of CO₂ produced during the reaction.

Thus, the volume of CO₂ formed = 20 mL.

### Step 5: Determine the volume of unreacted O₂
From the total volume of gas after the explosion (90 mL), we can calculate the volume of unreacted O₂:
- Volume of gas after explosion = Volume of unreacted O₂ + Volume of CO₂
- 90 mL = Volume of unreacted O₂ + 20 mL (CO₂)

Rearranging gives:
- Volume of unreacted O₂ = 90 mL - 20 mL = 70 mL

### Final Answer
The volume of unreacted O₂ is **70 mL**.

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