A crystalline solid dissolves in walur lo give colourless solution. On addition of aqueous silver nitrate, a yellow precipitate is formed. The solid is most likely to be

Options

A. copper (II) oxide
B. sodium chloride
C. sodium iodide
D. zinc sulphate..

Answer

To solve the question, we need to identify the crystalline solid based on the information provided about its behavior in water and its reaction with silver nitrate.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- The crystalline solid dissolves in water to give a colorless solution.
- Upon adding aqueous silver nitrate (AgNO3), a yellow precipitate is formed.

2. **Analyzing the Colorless Solution**:
- A colorless solution indicates that the solid is likely a salt that dissociates into ions without imparting any color. Common examples include sodium chloride (NaCl), potassium bromide (KBr), and sodium iodide (NaI).

3. **Identifying the Reaction with Silver Nitrate**:
- The formation of a yellow precipitate upon the addition of silver nitrate suggests that the solid contains an anion that reacts with silver ions (Ag+) to form a colored precipitate.
- Silver iodide (AgI) is known to form a yellow precipitate when silver nitrate is added to a solution containing iodide ions (I-).

4. **Testing Possible Solids**:
- Since we are looking for a solid that dissolves in water to yield a colorless solution and produces a yellow precipitate with AgNO3, we can consider sodium iodide (NaI).
- When NaI dissolves in water, it dissociates into Na+ and I- ions, resulting in a colorless solution.

5. **Confirming the Reaction**:
- When we add AgNO3 to the colorless solution of NaI, the iodide ions (I-) react with silver ions (Ag+) to form silver iodide (AgI):
\[ \text{NaI (aq)} + \text{AgNO}_3 \text{(aq)} \rightarrow \text{AgI (s)} + \text{NaNO}_3 \text{(aq)} \]
- Silver iodide (AgI) is a yellow precipitate, confirming our identification.

6. **Conclusion**:
- Based on the observations and the reactions, the crystalline solid is most likely sodium iodide (NaI).

### Final Answer:
The solid is most likely **sodium iodide (NaI)**.