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 (a) 32 (b) 64 (c) 128 (d) 488

Step-by-Step Solution

Answer

The molar mass of the substance is 32 g/mol.

Step 1: Determine Change in Vapour Pressure

The change in vapour pressure when the substance is added can be calculated by subtracting the vapour pressure of the solution from the vapour pressure of pure acetone. So, the change is \(185\, \text{torr} - 183\, \text{torr} = 2\, \text{torr}\).

Step 2: Calculate Relative Lowering of Vapour Pressure

The relative lowering of vapour pressure is given by \(\frac{\Delta P}{P_0} = \frac{2}{185}\), where \(\Delta P\) is the change in vapour pressure, and \(P_0\) is the original vapour pressure of acetone.

Step 3: Apply the Formula for Molar Mass

Using the formula for the relative lowering of vapour pressure for solutions, we have \(\frac{\Delta P}{P_0} = \frac{n_2}{n_1 + n_2}\approx \frac{n_2}{n_1}\), since \(n_2\) (moles of solute) is much smaller than \(n_1\) (moles of solvent). Here, \(n_1 = \frac{100}{58} \approx 1.724\), as the molar mass of acetone is \(58 \text{ g/mol}\).

Step 4: Calculate Moles of Solute

The relative lowering formula becomes \(\frac{2}{185} = \frac{1.2/M}{1.724}\), where \(M\) is the molar mass of the solute. Simplifying, \(\frac{1.2}{M} = \frac{2 \times 1.724}{185}\).

Step 5: Solve for M

Rearrange and solve \(M = \frac{1.2 \times 185}{2 \times 1.724}\), which calculates to \(M \approx 32.06\). Thus, the closest molar mass option is \(32\, \text{g/mol}\).

Key Concepts

Relative Lowering of Vapour Pressure

The concept of relative lowering of vapour pressure is essential to understanding how solutions behave, especially when a solute is added to a solvent. When a non-volatile solute is dissolved in a solvent, the vapour pressure of the resulting solution decreases. This decrease is due to the fact that some of the solvent molecules at the surface are replaced by solute molecules, which do not evaporate. Thus, the overall number of solvent molecules that can escape into the vapour phase reduces. Relative lowering of vapour pressure can be expressed mathematically as:

• \( \frac{\Delta P}{P_0} = \frac{P_0 - P_s}{P_0} \)

where \( \Delta P \) is the change in vapour pressure, \( P_0 \) is the vapour pressure of the pure solvent, and \( P_s \) is the vapour pressure of the solution. For our specific problem, the vapour pressure dropped from 185 torr to 183 torr upon adding the solute. This gives a relative lowering of vapour pressure as \( \frac{2}{185} \). Understanding this concept is crucial for calculating the molar mass of the solute in further steps.

Molar Mass Calculation

Calculating the molar mass of a solute in a solution involves using the concept of relative lowering of vapour pressure. From the previous concept, we learned how the vapour pressure changes. Now, using this change, the moles of solute can be correlated to the lost vapour pressure. The formula for this is:

• \( \frac{\Delta P}{P_0} = \frac{n_2}{n_1} \)

Here, \( n_2 \) and \( n_1 \) are the moles of solute and solvent, respectively. Given that the molar mass of acetone, the solvent, is 58 g/mol, we can find the moles of solvent \( n_1 \) as:

• \( n_1 = \frac{100}{58} \approx 1.724 \)

Substituting into the relative lowering equation, we solve for the molar mass \( M \) of the solute:

• \( M = \frac{1.2 \times 185}{2 \times 1.724} \)

The outcome is approximately 32 g/mol, which matches close to option (a) 32 g/mol in the original choices.

Non-volatile Solute

A non-volatile solute is a substance that does not readily evaporate into the gas phase. This property is particularly useful in solutions where the goal is to reduce the vapour pressure without losing solute to evaporation. In our acetone example, the given solute is non-volatile, meaning it has little to no tendency to become a gas under the given conditions. When added to acetone, it stays in the liquid phase and replaces some of the acetone molecules that would otherwise contribute to vapour pressure. This results in a lowered vapour pressure for the solution compared to the pure solvent. Knowing about non-volatile solutes is essential for understanding colligative properties such as vapour pressure lowering, boiling point elevation, and freezing point depression. Since these properties depend on the number of solute particles rather than their identity, they are crucial in calculating attributes like molar mass. By grasping how non-volatile solutes function, students can better predict how solutions behave in real-world applications, such as antifreeze in car radiators or in processes like distillation.