A chemist opened a cupboard to find four bottles containing water solutions, each of which has lost its label. Bottles 1,2,3 contained colourless solutions, Whilst bottle 4 contained a blue solution. The labels from the bottles were lying scattered ont he floor of the cupboard.
They were
copper (II) sulphate
sodium carbonate
lead nitrate
hydrochloric acid
By mixing sammples of teh contents of the bottles, in pairs, the chemist made the following observations:
(i) Bottle 1 + Bottle 2 white precipitate
(ii) Bottle 1 + bottle 3 white precipitate
(iii) Bottle 1 + Bottle 4 white precipitate
(iv) Bottle 2 +Bottle 3 colourless gas evolved
(v) Bottle 2+ Bottle 4 no visible reaction
(vi) bottle 3 +bottle 4 blue precipitate
Q. Nature of gas evolved in observation (iv) is:

Options

A. Acidic
B. Neutral
C. Basic
D. Amphoteric

Explanation

To determine the nature of the gas evolved in observation (iv) when mixing Bottle 2 and Bottle 3, we will analyze the information step by step.

### Step 1: Identify the contents of the bottles based on observations.

1. **Observation (i)**: Bottle 1 + Bottle 2 → White precipitate
2. **Observation (ii)**: Bottle 1 + Bottle 3 → White precipitate
3. **Observation (iii)**: Bottle 1 + Bottle 4 → White precipitate
4. **Observation (iv)**: Bottle 2 + Bottle 3 → Colourless gas evolved
5. **Observation (v)**: Bottle 2 + Bottle 4 → No visible reaction
6. **Observation (vi)**: Bottle 3 + Bottle 4 → Blue precipitate

### Step 2: Analyze the observations.

- **Bottles 1, 2, and 3** contain colorless solutions, while **Bottle 4** contains a blue solution, which we can identify as **Copper (II) Sulphate (CuSO₄)**.
- Since Bottle 4 is blue, it must be Copper (II) Sulphate.

### Step 3: Determine the contents of the other bottles.

- **Observation (vi)**: Bottle 3 + Bottle 4 → Blue precipitate. This indicates that Bottle 3 must contain a substance that reacts with Copper (II) Sulphate to form a blue precipitate. The only colorless solution that can react to form a precipitate with Copper (II) Sulphate is **Sodium Carbonate (Na₂CO₃)**, which is likely in Bottle 2.

- Therefore, we can conclude:
- **Bottle 1**: Lead Nitrate (Pb(NO₃)₂)
- **Bottle 2**: Sodium Carbonate (Na₂CO₃)
- **Bottle 3**: Hydrochloric Acid (HCl)
- **Bottle 4**: Copper (II) Sulphate (CuSO₄)

### Step 4: Analyze Observation (iv).

- **Observation (iv)**: Bottle 2 (Sodium Carbonate) + Bottle 3 (Hydrochloric Acid) → Colourless gas evolved.
- The reaction between Sodium Carbonate and Hydrochloric Acid is:
\[ \text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{CO}_2 + \text{H}_2\text{O} \]
- From this reaction, we see that Carbon Dioxide (CO₂) is produced, which is a colourless gas.

### Conclusion:

The nature of the gas evolved in observation (iv) is **Carbon Dioxide (CO₂)**.

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