Problem 167 - A Complete Resource Book in Chemistry for JEE Main

Question

If sodium sulphate is considered to be completely dissociated into cations and anions in aqueous solution, the change in freezing point of water \(\left(\Delta \mathrm{T}_{f}\right)\), when \(0.01 \mathrm{~mol}\) of sodium sulphate is dissolved in \(1 \mathrm{Kg}\) of water, is \(\left(\mathrm{K}_{f}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)\) (a) \(0.0372 \mathrm{~K}\) (b) \(0.0558 \mathrm{~K}\) (c) \(0.0744 \mathrm{~K}\) (d) \(0.0186 \mathrm{~K}\)

Step-by-Step Solution

Answer

The change in freezing point is 0.0558 K, option (b).

Step 1: Understanding Complete Dissociation

Sodium sulphate (Na₂SO₄) dissociates completely in water to form 2 sodium ions (Na⁺) and 1 sulphate ion (SO₄²⁻). This results in a total of 3 ions per formula unit of Na₂SO₄.

Step 2: Determining the Van't Hoff Factor

The Van't Hoff factor, denoted as 'i', is the number of particles the solute breaks into in solution. For Na₂SO₄, it completely dissociates into 3 ions, so i = 3.

Step 3: Applying the Freezing Point Depression Formula

The formula for freezing point depression is given by:\[ \Delta T_f = i \cdot K_f \cdot m \]where \(\Delta T_f\) is the depression in freezing point, \(i\) is the van't Hoff factor, \(K_f\) is the cryoscopic constant, and \(m\) is the molality of the solution.

Step 4: Calculating Molality

Since 0.01 mol of sodium sulphate is dissolved in 1 kg of water, the molality (m) is 0.01 mol/kg.

Step 5: Calculating the Change in Freezing Point

Substitute values into the freezing point depression formula:\[ \Delta T_f = 3 \times 1.86 \times 0.01 = 0.0558 \text{ K} \]Thus, the change in freezing point is \(0.0558 \text{ K}\).

Key Concepts

Van't Hoff Factor

The Van't Hoff factor is a key concept when studying the colligative properties of solutions. It represents the degree to which a solute dissociates in a solution.

• In simple terms, it tells us how many particles a compound will form when dissolved in a solvent.
• For example, if a compound doesn't dissociate, like sugar in water, the Van't Hoff factor (\(i\)) is 1.
• In the case of sodium sulphate (\(\text{Na}_2\text{SO}_4\)), it breaks down into\(3\) ions: two sodium ions (\(\text{Na}^+\)) and one sulphate ion (\(\text{SO}_4^{2-}\)).
• Thus, for sodium sulphate, the Van't Hoff factor is 3.

Sodium sulphate provides an excellent example of how the Van't Hoff factor can exceed 1 when complete dissociation occurs, impacting the colligative properties of the solution like freezing point depression.

Sodium Sulphate Dissociation

Understanding the dissociation of sodium sulphate is pivotal in aqueous solution chemistry. When sodium sulphate dissolves:

• It separates completely into its constituent ions: two sodium ions (\(\text{Na}^+\)) and one sulphate ion (\(\text{SO}_4^{2-}\)).
• This complete dissociation contributes to the presence of three particles per formula unit.

Dissociation is an essential phenomenon in chemistry, affecting the physical and chemical properties of the solution. In our example, the dissociation of sodium sulphate in water is considered complete, which simplifies the calculation of colligative properties like the freezing point depression.

Molality Calculation

Calculating molality is an important step when dealing with solutions. It is defined as the number of moles of solute per kilogram of solvent.

• In our problem, we have\(0.01\) moles of sodium sulphate dissolved in\(1\) kg of water.
• This gives us a simple molality (\(m\)) of \(0.01\text{ mol/kg}\).

Having the molality calculated accurately helps in applying the correct values in the freezing point depression formula, which is crucial for finding the change in freezing point. Knowing the molality also allows one to understand how the concentration of a solution can affect its properties.

Aqueous Solution Chemistry

Aqueous solution chemistry involves the study of solutes dissolved in water, a crucial aspect in numerous scientific and industrial processes.

• Water, being a polar solvent, effectively dissolves many ionic compounds like sodium sulphate.
• Once dissolved, these compounds dissociate into ions, which then interact with water molecules.

In our example, the dissociation of sodium sulphate into sodium and sulphate ions is illustrated, demonstrating the principle of aqueous chemistry in action. These interactions in solution lead to various colligative properties, such as freezing point depression, boiling point elevation, and osmotic pressure. Understanding the basics of aqueous solution chemistry is fundamental to predicting and manipulating these properties.