If you had equal masses of the substances in the following pairs of compounds, which of the two would contain the greater number of ions? (a) NaBr or KCl (b) NaCl or MgCl2 (c) CrCl3 or Na2S

To determine which compound would contain a greater number of ions when equal masses are considered, we need to look at the following factors:

1. **Molar Mass:** The molar mass of the compounds allows us to determine how many moles of each compound are present in an equal mass.
2. **Number of Ions per Formula Unit:** The number of ions generated by each formula unit of the compound when it dissociates in solution.

Let's evaluate each pair:

### (a) **NaBr or KCl**

- **NaBr (Sodium Bromide):**
  - Molar Mass: \( \text{Na} = 23 \, \text{g/mol}, \, \text{Br} = 79.9 \, \text{g/mol} \)
  - Molar Mass of NaBr: \( 23 + 79.9 = 102.9 \, \text{g/mol} \)
  - Dissociates into: \( \text{Na}^+ + \text{Br}^- \) (2 ions)

- **KCl (Potassium Chloride):**
  - Molar Mass: \( \text{K} = 39.1 \, \text{g/mol}, \, \text{Cl} = 35.45 \, \text{g/mol} \)
  - Molar Mass of KCl: \( 39.1 + 35.45 = 74.55 \, \text{g/mol} \)
  - Dissociates into: \( \text{K}^+ + \text{Cl}^- \) (2 ions)

Since the **molar mass of KCl is smaller**, you would have more moles of KCl for the same mass compared to NaBr. Both dissociate into 2 ions, so **KCl** would contain more ions.

### (b) **NaCl or MgCl2**

- **NaCl (Sodium Chloride):**
  - Molar Mass of NaCl: \( 23 + 35.45 = 58.45 \, \text{g/mol} \)
  - Dissociates into: \( \text{Na}^+ + \text{Cl}^- \) (2 ions)

- **MgCl2 (Magnesium Chloride):**
  - Molar Mass: \( \text{Mg} = 24.3 \, \text{g/mol}, \, \text{Cl} = 35.45 \, \text{g/mol} \)
  - Molar Mass of MgCl2: \( 24.3 + 2(35.45) = 95.2 \, \text{g/mol} \)
  - Dissociates into: \( \text{Mg}^{2+} + 2\text{Cl}^- \) (3 ions)

Even though **NaCl has a smaller molar mass**, **MgCl2 dissociates into 3 ions per formula unit**, while NaCl only produces 2 ions. Therefore, **MgCl2** would contain more ions for the same mass.

### (c) **CrCl3 or Na2S**

- **CrCl3 (Chromium(III) Chloride):**
  - Molar Mass: \( \text{Cr} = 52 \, \text{g/mol}, \, \text{Cl} = 35.45 \, \text{g/mol} \)
  - Molar Mass of CrCl3: \( 52 + 3(35.45) = 158.35 \, \text{g/mol} \)
  - Dissociates into: \( \text{Cr}^{3+} + 3\text{Cl}^- \) (4 ions)

- **Na2S (Sodium Sulfide):**
  - Molar Mass: \( 2(23) + 32.07 = 78.07 \, \text{g/mol} \)
  - Dissociates into: \( 2\text{Na}^+ + \text{S}^{2-} \) (3 ions)

Since **Na2S has a much smaller molar mass** than CrCl3, you'd get more moles of Na2S for the same mass. However, **CrCl3 produces more ions per formula unit (4 ions compared to 3 ions for Na2S)**. This makes it more likely that **CrCl3** would contain more ions.

### Summary:
- (a) **KCl** would have more ions.
- (b) **MgCl2** would have more ions.
- (c) **CrCl3** would have more ions.

Would you like to explore the detailed steps for any of these comparisons? Here are 5 related questions:

1. How does the dissociation of polyatomic ions affect ion count in a solution?
2. Why do smaller molar masses lead to more moles in equal masses of compounds?
3. How would the presence of hydration in salts affect the ion count?
4. How can you calculate the total number of ions in a given mass of an ionic compound?
5. What role does ionic charge play in solubility and dissociation?

**Tip:** Always check how many ions each formula unit of a compound produces upon dissociation—this can greatly affect the total ion count!

Explore how to determine which compound contains the greater number of ions when equal masses of NaBr, KCl, NaCl, MgCl2, CrCl3, and Na2S are compared. This problem involves key stoichiometric concepts, molar mass calculations, and ionic dissociation. Suitable for Grades 10-12.

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