Question

A $0.0020\, M$ aqueous solution of an ionic compound $Co \left( NH _{3}\right)_{5}\left( NO _{2}\right) Cl$ freezes at $-0.00732^{\circ} C$. Number of moles of ions which 1 mole of ionic compound produces on being dissolved in water will be : $\left( k _{ f }=1.86^{\circ} C / m \right)-$

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The Correct Option is B

To determine the number of moles of ions that one mole of the ionic compound Co \left( NH _{3}\right)_{5}\left( NO _{2}\right) Cl produces on being dissolved in water, we can use the concept of depression in freezing point.

The depression in freezing point is calculated using the formula:

\Delta T_f = i \cdot k_f \cdot m

Where:

• \Delta T_f is the depression in freezing point.
• i is the van't Hoff factor, which represents the number of moles of particles the compound produces in solution.
• k_f is the cryoscopic constant, provided as 1.86^\circ C / m.
• m is the molality of the solution.

Given data:

• Depression in freezing point, \Delta T_f = 0 - (-0.00732) = 0.00732^\circ C
• Molality, m = 0.0020\, m
• Cryoscopic constant, k_f = 1.86^\circ C / m

Substituting the values into the formula:

0.00732 = i \cdot 1.86 \cdot 0.0020

Simplifying for i:

i = \frac{0.00732}{1.86 \times 0.0020} = \frac{0.00732}{0.00372} \approx 1.97

The calculated van't Hoff factor, i, is approximately 2, indicating that the ionic compound dissociates into 2 ions in solution.

The possible dissociation of Co \left( NH _{3}\right)_{5}\left( NO _{2}\right) Cl can be represented as:

[Co(NH_3)_5(NO_2)]Cl \rightarrow [Co(NH_3)_5(NO_2)]^+ + Cl^−

This dissociation shows that the compound dissociates into two ions:

• [Co(NH_3)_5(NO_2)]^+
• Cl^−

Thus, the correct answer is 2 moles of ions are produced.