Question

The depression in freezing point of \(0.01 \mathrm{M}\) aqueous solution of urea, sodium chloride and sodium sulphate is in the ratio (a) \(1: 1: 1\) (b) \(1: 2: 3\) (c) \(1: 2: 4\) (d) \(2: 2: 3\)

Short answer

The answer is (b) 1:2:3.

Step-by-step solution

Understand Depression in Freezing Point

Depression in freezing point (ΔTf) is calculated using the formula \(ΔT_f = i imes K_f imes m\), where \(i\) is the van't Hoff factor, \(K_f\) is the cryoscopic constant, and \(m\) is the molality of the solution. Here, since we are comparing similar concentrations of substances, molality \(m\) and \(K_f\) can be considered constant, and the effect on freezing point depression depends solely on the van't Hoff factor \(i\).

Identify the Van't Hoff Factor

For urea, which does not dissociate, \(i = 1\). Sodium chloride (NaCl) dissociates into two ions (Na⁺ and Cl⁻), so \(i = 2\). Sodium sulphate (Na2SO4) dissociates into three ions (2 Na⁺ and SO4²⁻), so \(i = 3\).

Compare the Depression in Freezing Points

The depression in freezing point is proportional to \(i\). For urea (i=1), sodium chloride (i=2), and sodium sulphate (i=3), the corresponding freezing point depression ratios follow \(ΔT_f\): 1:2:3.

Match with Given Options

The calculated depression ratio of 1:2:3 matches option (b), which states the ratio as 1:2:3. Therefore, the correct answer is option (b).

Key concepts

Van't Hoff Factor

The Van’t Hoff factor, often denoted by the letter \(i\), is a dimensionless number used in chemistry to account for the effect of solutes on colligative properties like freezing point depression. In simple terms, it tells us how many particles a solute forms in a solution. For non-electrolytes like urea, which do not dissociate in a solution, \(i\) is equal to 1. This means they contribute one part per molecule to the colligative properties.

For ionic compounds, which dissociate into ions, \(i\) corresponds to the number of ions formed. Taking sodium chloride (NaCl) as an example, it dissociates into sodium ions (Na\(^+\)) and chloride ions (Cl\(^-\)), resulting in \(i = 2\). Sodium sulfate (Na\(_2\)SO\(_4\)) dissociates into two sodium ions and one sulfate ion (SO\(_4\)\(^{2-}\)), yielding \(i = 3\).

Understanding the van’t Hoff factor is crucial when calculating various properties of solutions because it influences how we predict changes in a solution’s physical properties, particularly those affected by the number of solute particles, like freezing point depression.

Cryoscopic Constant

The cryoscopic constant, represented as \(K_f\), is a property of the solvent and indicates how its freezing point is lowered per molal concentration of a non-volatile solute. It is expressed in degrees Celsius per molal (6C·kg/mol).

Each solvent has a specific \(K_f\) value. For example, water, which is common in freezing point depression studies, has a distinct cryoscopic constant that determines how much its freezing point will decrease for a given amount of solute.

The cryoscopic constant emerges in the freezing point depression equation \(ΔT_f = i \times K_f \times m\). This formula shows how the freezing point drop \(ΔT_f\) is proportional to \(i\), the van't Hoff factor, \(K_f\), and \(m\), the solution's molality. This relationship makes \(K_f\) crucial in accurately predicting changes in the freezing point, contributing to a better understanding of solution behaviors when different solutes are present.

Molality in Solutions

Molality, denoted as \(m\), is a way to express the concentration of a solution. Unlike molarity, which is volume-based, molality is defined as moles of solute per kilogram of solvent. This makes it unaffected by temperature changes, offering a more stable concentration measure in varying conditions.

Molality is especially useful when dealing with colligative properties like freezing point depression. The formula \(ΔT_f = i \times K_f \times m\) directly incorporates molality to determine how a solute affects the freezing point of a solvent.

Since molality does not depend on temperature, it is particularly reliable for studying physical properties and comparing different solutions. During experiments, knowing the molality helps accurately assess the quantity of a solute in a given amount of solvent, thereby allowing precise calculations of any resulting changes in physical properties like the freezing point.