A 0.0020 m aqueous solution of an ionic compound `Co(NH_(3))_(5) (NO_(2))Cl` freezes at `-0.00744^(@)C`. The number of moles of ions which 1 mole of ionic compound produces on being dissolve in water is `(K_(f)=-1.86^(@)C//m)`

Explanation

To solve the problem, we need to determine the number of moles of ions produced when 1 mole of the ionic compound \( Co(NH_3)_5(NO_2)Cl \) is dissolved in water. We will use the freezing point depression formula.

### Step-by-Step Solution:

1. **Identify the given values:**
- Molality of the solution, \( m = 0.0020 \, \text{m} \)
- Freezing point depression, \( \Delta T_f = -0.00744^\circ C \)
- Freezing point depression constant for water, \( K_f = -1.86^\circ C/m \)

2. **Use the freezing point depression formula:**
\[ \Delta T_f = i \cdot K_f \cdot m \]
where \( i \) is the van 't Hoff factor (the number of particles the solute dissociates into).

3. **Rearranging the formula to find \( i \):**
\[ i = \frac{\Delta T_f}{K_f \cdot m} \]

4. **Substituting the known values into the equation:**
\[ i = \frac{-0.00744}{-1.86 \cdot 0.0020} \]

5. **Calculating the denominator:**
\[ -1.86 \cdot 0.0020 = -0.00372 \]

6. **Now substituting back to find \( i \):**
\[ i = \frac{-0.00744}{-0.00372} = 2.0 \]

7. **Conclusion:**
The number of moles of ions produced by 1 mole of the ionic compound \( Co(NH_3)_5(NO_2)Cl \) is \( i = 2.0 \).

### Final Answer:
The number of moles of ions produced when 1 mole of \( Co(NH_3)_5(NO_2)Cl \) is dissolved in water is **2 moles**.

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