Problem 173 - A Complete Resource Book in Chemistry for JEE Main

Question

The vapour pressure of acetone at \(20^{\circ} \mathrm{C}\) is 185 torr. When \(1.2 \mathrm{~g}\) of a non-volatile substance was dissolved in \(100 \mathrm{~g}\) of acetone at \(20^{\circ} \mathrm{C}\), its vapour pressure was 183 torr. The molar mass \(\left(\mathrm{g} \mathrm{mol}^{-1}\right)\) of the substance is [2015] (a) 32 (b) 64 (c) 128 (d) 488

Step-by-Step Solution

Answer

The molar mass of the substance is 64 g/mol (Option b).

Step 1: Understanding the Problem

We are given the initial vapor pressure of pure acetone, which is 185 torr, and the vapor pressure of acetone with a non-volatile solute, which is 183 torr. We need to find the molar mass of the solute.

Step 2: Applying Raoult's Law

Raoult's Law states that the vapor pressure of a solution, \( P_{solution} \), is equal to the mole fraction of the solvent \( X_{solvent} \) times the vapor pressure of the pure solvent \( P^0_{solvent} \). The change in vapor pressure \( \Delta P = P^0_{solvent} - P_{solution} \).

Step 3: Calculating Mole Fraction

From Raoult's Law, we have: \[P_{solution} = X_{acetone} \times P_{acetone}^0\]Re-arranging gives:\[X_{acetone} = \frac{P_{solution}}{P_{acetone}^0} = \frac{183}{185}\]

Step 4: Calculating Moles of Acetone

The mole fraction equation is \[X_{acetone} = \frac{n_{acetone}}{n_{acetone} + n_{solute}}\]First, calculate moles of acetone:\[ \text{Molar mass of acetone} = 58 \, \text{g/mol}\]\[n_{acetone} = \frac{100}{58}\]

Step 5: Setting up the Equation for Solute

Using the calculated \(X_{acetone}\) and rearranging the mole fraction formula:\[\frac{183}{185} = \frac{n_{acetone}}{n_{acetone} + \frac{1.2}{M_{solute}}}\]Simplifying gives:\[\frac{1.0}{1.0 + \frac{1.2}{M_{solute}}}\]

Step 6: Solving for Molar Mass of the Solute

Solve for \(M_{solute}\):Use the equation from the previous step and solve it for \(M_{solute}\).\[\text{By solving, } M_{solute} = 64 \, \text{g/mol}\]

Step 7: Selecting the Correct Answer

Among the options provided (a) 32, (b) 64, (c) 128, (d) 488, the molar mass of the non-volatile substance is 64 g/mol, which is option (b).

Key Concepts

Vapor Pressure

Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid or solid phase at a given temperature. It's a crucial concept as it helps us understand how different substances interact when mixed. When a non-volatile solute is added to a solvent, it reduces the surface area accessible for the solvent molecules to escape as vapor, leading to a decreased vapor pressure. In our exercise, the initial vapor pressure of pure acetone is 185 torr at 20°C. By adding a non-volatile solute, the vapor pressure drops to 183 torr, highlighting the change caused by solute-solvent interaction.

Mole Fraction

The mole fraction is a way of expressing the concentration of a component in a mixture, calculated as the ratio of moles of that component to the total number of moles in the mixture. It's a dimensionless number, meaning it has no units, and ranges from 0 to 1. The equation for the mole fraction of acetone in the solution is derived from Raoult's Law: \( X_{acetone} = \frac{P_{solution}}{P_{acetone}^0} = \frac{183}{185} \). This step is essential for finding the concentration of the acetone, which further helps us in determining the amount of the solute added.

Molar Mass Calculation

Calculating the molar mass involves finding the mass of one mole of a substance. In this exercise, we need to find the molar mass of the non-volatile solute dissolved in acetone. After calculating the mole fraction, we use the formula: \( X_{acetone} = \frac{n_{acetone}}{n_{acetone} + n_{solute}} \), where \( n_{acetone} \), the moles of acetone, is given as \( \frac{100}{58} \). We solve the equation: \( \frac{183}{185} = \frac{n_{acetone}}{n_{acetone} + \frac{1.2}{M_{solute}}} \). By rearranging and simplifying, we find the molar mass of the solute to be 64 g/mol.

Non-Volatile Solute

A non-volatile solute is a substance that has a negligible vapor pressure compared to its solvent. This means it doesn't evaporate easily. When dissolved in a solvent, it causes a decrease in the solvent's vapor pressure, a crucial point in understanding Raoult's Law. Non-volatile solutes are often used in solutions to elevate boiling points or lower freezing points. In this exercise, the non-volatile solute is responsible for the reduction in acetone’s vapor pressure, from 185 torr to 183 torr, facilitating the calculation of the solute’s molar mass.