Problem 90 - A Complete Resource Book in Chemistry for JEE Main

Question

On mixing \(3 \mathrm{~g}\) of non-volatile solute in \(200 \mathrm{~mL}\) of water its boiling point \(\left(100^{\circ} \mathrm{C}\right)\) becomes \(100.52^{\circ} \mathrm{C}\). If \(\mathrm{K}_{b}\) for water is \(0.6 \mathrm{~K} / \mathrm{m}\) then molecular weight of the solute is (a) \(10.5 \mathrm{~g} \mathrm{~mol}^{-1}\) (b) \(12.6 \mathrm{~g} \mathrm{~mol}^{-1}\) (c) \(15.7 \mathrm{~g} \mathrm{~mol}^{-1}\) (d) \(17.3 \mathrm{~g} \mathrm{~mol}^{-1}\)

Step-by-Step Solution

Answer

The molecular weight of the solute is about 17.3 g/mol.

Step 1: Calculate the Boiling Point Elevation

The boiling point elevation, \( \Delta T_b \), is the difference between the boiling point of the solution and the boiling point of the pure solvent. Here, \( \Delta T_b = 100.52^{\circ} \mathrm{C} - 100^{\circ} \mathrm{C} = 0.52^{\circ} \mathrm{C} \).

Step 2: Use the Boiling Point Elevation Formula

Boiling point elevation is given by the formula \( \Delta T_b = i \cdot K_b \cdot m \), where \( i \) is the van't Hoff factor (which is 1 for non-electrolytes), \( K_b \) is the ebullioscopic constant, and \( m \) is the molality of the solution. We have \( 0.52 = 0.6 \times m \). Solve for \( m \): \[ m = \frac{0.52}{0.6} = 0.8667 \text{ mol/kg} \]

Step 3: Convert Volume of Solvent to Mass

Assuming the density of water is \( 1 \mathrm{~g/mL} \), the mass of \( 200 \mathrm{~mL} \) of water is \( 200 \mathrm{~g} \), which is also \( 0.2 \mathrm{~kg} \).

Step 4: Calculate the Moles of Solute

Molality \( m \) is defined as moles of solute per kilogram of solvent. We know \( m = 0.8667 \mathrm{~mol/kg} \) and the mass of the solvent is \( 0.2 \mathrm{~kg} \). Therefore, number of moles of solute \( n \) is:\[ n = 0.8667 \times 0.2 = 0.17334 \text{ mol} \]

Step 5: Calculate Molecular Weight

The molecular weight \( M \) of the solute can be calculated using the mass of the solute and the number of moles obtained. The mass of the solute is \( 3 \mathrm{~g} \). Therefore, \[ M = \frac{3 \mathrm{~g}}{0.17334 \mathrm{~mol}} \approx 17.3 \mathrm{~g/mol} \]

Step 6: Choose the Correct Option

Compare the calculated molecular weight with the given options. The molecular weight of the solute is closest to option \((d)\).

Key Concepts

Molality

Molality is a measure of the concentration of a solute in a solvent. However, unlike molarity, it does not depend on the volume, which can change with temperature. Molality is defined as the number of moles of solute per kilogram of solvent.

It is important in scenarios where temperature changes, such as boiling or freezing, affect measurements because:

• Molality remains constant with temperature fluctuations since it is based on mass rather than volume.
• This attribute makes molality a reliable concentration unit for boiling point elevation calculations, as seen in the exercise.

Understanding molality helps in accurately determining changes in boiling or freezing points when a solute is dissolved in a solvent. This is crucial for analysis in chemistry and various industrial applications.

Ebullioscopic Constant

The ebullioscopic constant, denoted as \( K_b \), is a specific property of the solvent. It quantifies the boiling point elevation caused by the addition of a solute. It is used in the formula:\[\Delta T_b = i \cdot K_b \cdot m\]where \( i \) is the van't Hoff factor (1 for non-electrolytes as in this exercise), \( K_b \) is the ebullioscopic constant itself, and \( m \) represents molality.

This constant is unique to each solvent and must be determined experimentally. For example:

• Water has a known \( K_b \) value of 0.6 K/m.
• It allows us to calculate the elevation in boiling point when a solute is added.

Knowing the ebullioscopic constant is essential for practical applications. It assists chemists in predicting how a solution's boiling point will change with varying solute concentrations.

Molecular Weight Determination

Molecular weight, or molar mass, refers to the mass of one mole of a given chemical compound. It is critically important for stoichiometry and helps chemists understand the properties of compounds. In the exercise:

• The method used for calculating molecular weight involves measuring the effect of a solute on solvent properties, such as boiling point elevation.
• Knowing the weight of the solute and the moles of solute determined from molality, molecular weight can be accurately computed.

This technique, based on colligative properties, is practical for identifying unknown substances. It reveals how solute particles interact with the solvent molecules, pivotal for both academic research and industrial processes.

Non-volatile Solute

A non-volatile solute is one that does not easily vaporize. This characteristic is important when studying boiling point elevation, as is the case in the exercise.

Here's why non-volatile solutes matter:

• They increase the boiling point of solvents without evaporating into the gas phase.
• This quality ensures consistent results in boiling point elevation experiments because they stay in the liquid phase.

Non-volatile solutes are typically used to explore colligative properties, as they allow clear observation of changes in boiling or freezing points. This results in a more precise control over experiments concerning solution behaviors.