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. Colourless solution present in bottle-1 is:

Options

A. `CuSO_(4)`
B. `HCl`
C. `Pb(NO_(3))_(2)`
D. `Na_(2)CO_(3)`

Explanation

To determine the colorless solution present in bottle 1, we can analyze the observations made from mixing the solutions in pairs. Here’s a step-by-step solution:

### Step 1: Analyze the Observations
We have the following observations from mixing the bottles:
1. Bottle 1 + Bottle 2 → White precipitate
2. Bottle 1 + Bottle 3 → White precipitate
3. Bottle 1 + Bottle 4 → White precipitate
4. Bottle 2 + Bottle 3 → Colourless gas evolved
5. Bottle 2 + Bottle 4 → No visible reaction
6. Bottle 3 + Bottle 4 → Blue precipitate

### Step 2: Identify Possible Reactions
From the observations, we can deduce the following:
- The formation of a white precipitate in observations 1, 2, and 3 suggests that one of the solutions in bottle 1 must be a compound that can react with the others to form a precipitate.
- The white precipitate formed in observation 1 indicates that bottle 1 and bottle 2 contain lead nitrate (Pb(NO3)2) and hydrochloric acid (HCl), respectively, which react to form lead chloride (PbCl2), a white precipitate.

### Step 3: Confirm Bottle 1 Contains Lead Nitrate
Since bottle 1 reacts with bottle 2 to form a white precipitate, we can conclude that:
- Bottle 1 = Lead Nitrate (Pb(NO3)2)
- Bottle 2 = Hydrochloric Acid (HCl)

### Step 4: Analyze Other Observations
- In observation 4, the reaction between bottle 2 and bottle 3 produces a colorless gas, which is likely carbon dioxide (CO2). This suggests that bottle 3 is sodium carbonate (Na2CO3), as it reacts with hydrochloric acid to release CO2.
- In observation 5, the lack of reaction between bottle 2 and bottle 4 indicates that bottle 4 must be copper (II) sulfate (CuSO4), since it does not react with hydrochloric acid.
- Finally, observation 6 confirms that bottle 3 (sodium carbonate) and bottle 4 (copper (II) sulfate) react to form a blue precipitate of copper (II) carbonate (CuCO3).

### Conclusion
Based on the analysis, the colorless solution present in bottle 1 is **Lead Nitrate (Pb(NO3)2)**.