A chemist opened a cupboard and found four bottles containing water solutions, each of which had lost its label.Bottles 1,2,3 contanied colourless solution, while bottle 4 contained a blue solution.The labels from the bottles were lying scattered on the floor of the cupboard.They were:
copper (II) sulphate, Hydrochloric acid
lead nitrate , Sodium carbonate
By mixing samples of the contents of the bottles, in pairs , the chemist made the following observations :
Bottle 1 +Bottle 2 `to` White precipitate is formed.
Bottle 1 +Bottle 3 `to` White precipitate is formed.
Bottle 1 +Bottle 4 `to` White precipitate is formed.
Bottle 2 +Bottle 3 `to` Colourless and odourless gas is evolved.
Bottle 2 +Bottle 4 `to` No visible reaction is observed.
Bottle 3 +Bottle 4 `to` Blue precipitate is formed.
With the help of the above observations answer the following questions.
Bottle 3 contains :

Options

A. Copper (II) sulphate
B. Hydrochloric acid
C. Lead nitrate
D. Sodium carbonate

Explanation

To determine the contents of Bottle 3, we will analyze the observations made by the chemist step by step.

### Step-by-Step Solution:

1. **Identify the Solutions**:
- We know that Bottle 4 contains a blue solution, which is identified as copper (II) sulfate (CuSO₄).
- The remaining three bottles (1, 2, and 3) contain colorless solutions.

2. **Observation Analysis**:
- **Observation 1**: Mixing Bottle 1 and Bottle 2 results in a white precipitate.
- **Observation 2**: Mixing Bottle 1 and Bottle 3 also results in a white precipitate.
- **Observation 3**: Mixing Bottle 1 and Bottle 4 results in a white precipitate.
- **Observation 4**: Mixing Bottle 2 and Bottle 3 produces a colorless, odorless gas.
- **Observation 5**: Mixing Bottle 2 and Bottle 4 shows no visible reaction.
- **Observation 6**: Mixing Bottle 3 and Bottle 4 results in a blue precipitate.

3. **Determine Contents of Bottles**:
- From **Observation 1** (Bottle 1 + Bottle 2 = White precipitate), we suspect that Bottle 1 contains lead nitrate (Pb(NO₃)₂) and Bottle 2 contains hydrochloric acid (HCl), as these react to form lead(II) chloride (PbCl₂), which is a white precipitate.
- From **Observation 2** (Bottle 1 + Bottle 3 = White precipitate), if Bottle 1 is lead nitrate, then Bottle 3 must contain sodium carbonate (Na₂CO₃) because lead carbonate (PbCO₃) is also a white precipitate.
- **Observation 4** (Bottle 2 + Bottle 3 = Colorless gas evolved) indicates that when HCl reacts with sodium carbonate, carbon dioxide (CO₂) gas is released, which is colorless and odorless.
- **Observation 5** (Bottle 2 + Bottle 4 = No reaction) suggests that HCl does not react with copper sulfate, which is consistent.
- **Observation 6** (Bottle 3 + Bottle 4 = Blue precipitate) confirms that sodium carbonate reacts with copper sulfate to form copper carbonate (CuCO₃), which is a blue precipitate.

4. **Conclusion**:
- Based on the above deductions, we conclude that Bottle 3 contains sodium carbonate (Na₂CO₃).

### Final Answer:
**Bottle 3 contains: Sodium Carbonate (Na₂CO₃)**

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