Problem 119 - A Complete Resource Book in Chemistry for JEE Main

Question

The vapour pressure of a solution of \(5 \mathrm{~g}\) of non electrolyte in \(100 \mathrm{~g}\) of water at a particular temperature is \(2985 \mathrm{Nm}^{-2}\). The vapour pressure of pure water at that temperature is \(3000 \mathrm{Nm}^{-2}\). The molecular weight of the solute is (a) 180 (b) 90 (c) 270 (d) 360

Step-by-Step Solution

Answer

The molecular weight of the solute is 180 g/mol.

Step 1: Determine Given Quantities

The vapour pressure of the solution is given as 2985 Nm\(^{-2}\), and the vapour pressure of pure water is 3000 Nm\(^{-2}\). We have 5 g of solute and 100 g of water.

Step 2: Apply Raoult's Law for Vapour Pressure

Raoult's Law for vapour pressure states that the relative lowering of vapour pressure is equal to the mole fraction of the solute. It can be expressed as: \[ P^0 - P = \frac{P^0 \cdot n_2}{n_1 + n_2} \] where \( P^0 \) is the vapour pressure of pure solvent, \( P \) is the vapour pressure of the solution, \( n_1 \) is moles of solvent, and \( n_2 \) is moles of solute.

Step 3: Calculate Mole Fractions

We calculate the moles of the solvent (water) using its molar mass (18 g/mol). \[ n_1 = \frac{100}{18} \approx 5.56 \text{ mol} \] Let \( M \) be the molar mass of the solute, then moles of the solute \( n_2 \) is \( \frac{5}{M} \).

Step 4: Set Up Equation for Relative Lowering of Vapour Pressure

The relative lowering of vapour pressure can be written as: \[ \frac{3000 - 2985}{3000} = \frac{\frac{5}{M}}{5.56 + \frac{5}{M}} \] Simplifying the left side: \[ \frac{15}{3000} = 0.005 \]

Step 5: Solve for Molar Mass M

Substitute the values into the equation: \[ 0.005 = \frac{\frac{5}{M}}{5.56 + \frac{5}{M}} \] Cross-multiply to solve for \( M \): \[ 0.005 \cdot (5.56 + \frac{5}{M}) = \frac{5}{M} \] \[ 0.0278 + \frac{0.025}{M} = \frac{5}{M} \] \[ 0.0278 = \frac{5}{M} - \frac{0.025}{M} \] \[ 0.0278 = \frac{4.975}{M} \] \[ M = \frac{4.975}{0.0278} \] \[ M \approx 179.14 \] Rounding, we find \( M \approx 180 \text{ g/mol} \).

Step 6: Verify and Select Correct Option

The closest correct option to our calculated molecular weight is option (a) 180. Thus, the molecular weight of the solute is 180 g/mol.

Key Concepts

Raoult's Law

Raoult's Law deals with the vapour pressure of a solution, which is an important concept in chemistry. It states that the partial vapour pressure of each component in a solution is directly proportional to its mole fraction in the solution. The total vapour pressure of the solution is the sum of these partial pressures. Raoult's Law is represented mathematically for a solvent as:- \[ P = P^0 \times X_1 \]- where \( P^0 \) is the vapour pressure of the pure solvent, and \( X_1 \) is the mole fraction of the solvent.- In our exercise, the relative lowering of vapour pressure (\( P^0 - P \)) is related to the presence of a non-volatile solute, leading to a decrease in the solvent's vapour pressure.Understanding Raoult's Law helps to explain why adding a solute to a solvent lowers the vapour pressure of the solvent.

Molecular Weight Calculation

The molecular weight of a solute is crucial in many chemical calculations, as it allows scientists to convert between mass and moles. In the exercise, we used the relative lowering of vapour pressure to determine the solute's molecular weight. This involved setting up an equation using Raoult's Law and solving for the molar mass \( M \):- \[ 0.005 = \frac{\frac{5}{M}}{5.56 + \frac{5}{M}} \]- Solving this equation involved algebraic manipulation and cross-multiplying to isolate and solve for \( M \), resulting in approximately 180 g/mol.This process demonstrates the importance of understanding chemical equations and using them to find molecular weights for laboratory activities, such as creating solutions with precise concentrations.

Solute and Solvent

In a solution, the solute and solvent play distinct roles. The solvent is the substance present in a larger amount that dissolves the solute. Simply put, in any solution, the solute is the substance that is being dissolved. In our example, water is the solvent, and the non-electrolyte solute is what gets dissolved into it. - When you dissolve the solute into the solvent, it impacts the physicochemical properties such as vapour pressure, boiling point, and freezing point. - Understanding the nature of solute and solvent interactions is vital when preparing solutions in chemistry, helping to predict how the mixture will behave under different conditions.

Mole Fraction

The mole fraction is a way to express the concentrations of components in a mixture. It's defined as the ratio of the moles of one component to the total moles in the solution. It is a dimensionless quantity and does not depend on temperature. For the solvent in our example, the mole fraction is calculated by:- \[ X_1 = \frac{n_1}{n_1 + n_2} \]- Here \( n_1 \) is the moles of solvent and \( n_2 \) is the moles of solute.In the calculation step of the problem, this mole fraction was a key component in applying Raoult's Law, showing its importance in determining how much the vapour pressure is lowered by adding a solute.

Relative Lowering of Vapour Pressure

Relative lowering of vapour pressure refers to the decrease in vapour pressure when a non-volatile solute is added to a solvent. It quantifies this difference relative to the vapour pressure of the pure solvent, expressed as:- \[ \frac{P^0 - P}{P^0} \] - In the exercise, the vapour pressure of pure water (\( P^0 \)) is 3000 Nm\(^{-2}\), and with solute, it is 2985 Nm\(^{-2}\). Hence, the relative reduction is calculated as \( \frac{15}{3000} \).Understanding this concept is crucial because it shows how solutions behave differently from pure solvents and is key to many applications, like calculating boiling points and freezing points in solutions.