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. Chemical formula of white precipitate in observation (i) is:

Options

A. `CuCl_(2)`
B. `PbCl_(2)`
C. `PbCO_(3)`
D. `CuSO_(3)`

Explanation

To solve the problem, we need to analyze the observations made by the chemist when mixing the solutions in the bottles. Let's break down the information step by step.

### Step 1: Identify the Solutions in Each Bottle
We have four bottles with the following possible solutions:
- Bottle 1: Unknown
- Bottle 2: Unknown
- Bottle 3: Unknown
- Bottle 4: Blue solution (which must be Copper (II) Sulphate, CuSO₄)

### Step 2: Analyze the 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 → Colorless gas evolved
5. **Observation (v)**: Bottle 2 + Bottle 4 → No visible reaction
6. **Observation (vi)**: Bottle 3 + Bottle 4 → Blue precipitate

### Step 3: Deduce the Contents of Each Bottle
From the observations:
- **Observation (i)** and **(ii)** indicate that Bottle 1 must contain Lead Nitrate (Pb(NO₃)₂) since it forms a white precipitate with both Bottle 2 and Bottle 3.
- **Observation (iv)** suggests that Bottle 2 contains Hydrochloric Acid (HCl) because it reacts with Bottle 3 to produce a colorless gas (which is likely CO₂ from a carbonate).
- **Observation (vi)** indicates that Bottle 3 must contain Sodium Carbonate (Na₂CO₃) because it reacts with Copper (II) Sulphate to form a blue precipitate of Copper (II) Carbonate (CuCO₃).
- **Observation (v)** confirms that since there is no reaction between Bottle 2 (HCl) and Bottle 4 (CuSO₄), Bottle 4 must indeed be Copper (II) Sulphate.

### Step 4: Write the Chemical Reaction for Observation (i)
Now, we can confirm the chemical reaction for Observation (i):
- **Bottle 1 (Lead Nitrate)** + **Bottle 2 (Hydrochloric Acid)** → **Lead Chloride (PbCl₂)** (white precipitate)

The balanced chemical equation is:
\[ \text{Pb(NO}_3\text{)}_2 + 2\text{HCl} \rightarrow \text{PbCl}_2 \downarrow + 2\text{HNO}_3 \]

### Conclusion
The chemical formula of the white precipitate formed in Observation (i) is **Lead Chloride (PbCl₂)**.