`Na_(2)SO_(3),NaCl,Na_(2)C_(2)O_(4),Na_(2)HPO_(4),Na_(2)CrO_(4),NaNO_(2),CH_(3)CO_(2)Na` are separately treated with `AgNO_(3)` solution in how many cases is/are white ppt obtained ?

Explanation

To solve the question of how many white precipitates are formed when the given sodium salts are treated with silver nitrate (AgNO₃), we need to analyze each compound one by one.

### Step-by-Step Solution:

1. **Identify the Compounds**: The compounds given are:
- Na₂SO₃ (Sodium Sulfite)
- NaCl (Sodium Chloride)
- Na₂C₂O₄ (Sodium Oxalate)
- Na₂HPO₄ (Sodium Hydrogen Phosphate)
- Na₂CrO₄ (Sodium Chromate)
- NaNO₂ (Sodium Nitrite)
- CH₃CO₂Na (Sodium Acetate)

2. **Treat Each Compound with AgNO₃**:
- **Na₂SO₃**: When treated with AgNO₃, it forms Ag₂SO₃ (Silver Sulfite), which is a white precipitate.
- **NaCl**: When treated with AgNO₃, it forms AgCl (Silver Chloride), which is a white precipitate.
- **Na₂C₂O₄**: When treated with AgNO₃, it forms Ag₂C₂O₄ (Silver Oxalate), which is a white precipitate.
- **Na₂HPO₄**: When treated with AgNO₃, it does not form a white precipitate; it remains soluble.
- **Na₂CrO₄**: When treated with AgNO₃, it forms Ag₂CrO₄ (Silver Chromate), which is a brown-red precipitate, not white.
- **NaNO₂**: When treated with AgNO₃, it forms AgNO₂ (Silver Nitrite), which is a white precipitate.
- **CH₃CO₂Na**: When treated with AgNO₃, it forms CH₃CO₂Ag (Silver Acetate), which is a white precipitate.

3. **Count the Cases with White Precipitates**:
- Na₂SO₃ → Ag₂SO₃ (white)
- NaCl → AgCl (white)
- Na₂C₂O₄ → Ag₂C₂O₄ (white)
- Na₂HPO₄ → No precipitate (not white)
- Na₂CrO₄ → No precipitate (not white)
- NaNO₂ → AgNO₂ (white)
- CH₃CO₂Na → CH₃CO₂Ag (white)

4. **Total Count**: The total number of cases that produce white precipitates is:
- Na₂SO₃: 1
- NaCl: 1
- Na₂C₂O₄: 1
- NaNO₂: 1
- CH₃CO₂Na: 1

Thus, the total number of cases with white precipitates is **5**.

### Final Answer:
**5 cases produce white precipitates when treated with AgNO₃.**

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