Problem 83 - A Complete Resource Book in Chemistry for JEE Main

Question

If \(\mathrm{P}^{0}\) and \(\mathrm{P}\) are the vapour pressure of solvent and solution and \(\mathrm{n}_{1}\) and \(\mathrm{n}_{2}\) are the moles of solute and solvent respectively (a) \(\mathrm{P}^{0}=\mathrm{P}\left[\frac{\mathrm{n}_{1}}{\mathrm{n}_{1}+\mathrm{n}_{2}}\right]\) (b) \(\mathrm{P}^{0}=\mathrm{P}\left[\frac{\mathrm{n}_{2}}{\mathrm{n}_{1}+\mathrm{n}_{2}}\right]\) (c) \(\mathrm{P}=\mathrm{P}^{0}\left[\frac{\mathrm{n}_{2}}{\mathrm{n}_{1}+\mathrm{n}_{2}}\right]\) (d) \(\mathrm{P}^{0}=\mathrm{P} \times \mathrm{n}_{1}\)

Step-by-Step Solution

Answer

Correct option is (c): \(\mathrm{P} = \mathrm{P}^{0} \left[ \frac{\mathrm{n}_{2}}{\mathrm{n}_{1} + \mathrm{n}_{2}} \right]\).

Step 1: Understand Vapor Pressure

The vapor pressure of a solvent changes when a solute is added. Raoult's law explains this relationship. According to the law, the presence of a solute lowers the vapor pressure of the solvent.

Step 2: Define Given Symbols

We are given \(\mathrm{P}^{0}\) as the vapor pressure of the pure solvent, \(\mathrm{P}\) as the vapor pressure of the solution, \(\mathrm{n}_{1}\) as the moles of solute, and \(\mathrm{n}_{2}\) as the moles of solvent.

Step 3: Apply Raoult's Law

Raoult's law for the solution is given by \(\mathrm{P} = \mathrm{P}^{0} \times \left( \frac{\mathrm{n}_{2}}{\mathrm{n}_{1} + \mathrm{n}_{2}} \right)\). This equation represents the partial vapor pressure of the solvent in the solution.

Step 4: Analyze Each Option

We need to check which option matches the correct expression from Raoult's law:- (a) \(\mathrm{P}^{0} = \mathrm{P} \left[ \frac{\mathrm{n}_{1}}{\mathrm{n}_{1} + \mathrm{n}_{2}} \right]\) does not match Raoult's law.- (b) \(\mathrm{P}^{0} = \mathrm{P} \left[ \frac{\mathrm{n}_{2}}{\mathrm{n}_{1} + \mathrm{n}_{2}} \right]\) suggests partial pressure is expressed with \(\mathrm{P}\), not \(\mathrm{P}^{0}\).- (c) \(\mathrm{P} = \mathrm{P}^{0} \left[ \frac{\mathrm{n}_{2}}{\mathrm{n}_{1} + \mathrm{n}_{2}} \right]\) matches Raoult's law.- (d) \(\mathrm{P}^{0} = \mathrm{P} \times \mathrm{n}_{1}\) is not valid as it doesn't involve a mole fraction.

Key Concepts

Vapor Pressure

Vapor pressure is a key concept in understanding how liquids and gases behave in different conditions. Simply put, vapor pressure is the pressure exerted by the vapor that sits right above a liquid when the liquid and vapor are in equilibrium. This means that the rate at which the liquid evaporates into the gas phase is equal to the rate at which the gas condenses back into the liquid. Different liquids have different vapor pressures. This is because molecules in each liquid have different capabilities of escaping into the gas phase. For example, water has a moderate vapor pressure, whereas substances like alcohol, which evaporates more easily, have higher vapor pressures.
When a solute is added to a solvent, the vapor pressure of the resulting solution changes. This phenomenon is explained by Raoult's Law. Understanding vapor pressure helps us predict how solutions behave under various temperatures and concentrations. This concept becomes especially important when working with solutions in chemistry.

Solute and Solvent Interactions

The interaction between solute and solvent molecules plays a fundamental role in determining how a solution behaves. When a solute is added to a solvent, molecules from the solute interact with those of the solvent. This usually results in a lowering of the solvent's vapor pressure.

• The presence of a solute hinders the ability of solvent molecules to escape into the gas phase as easily as they would in a pure solvent.
• This interaction creates a solution that is different from the pure solvent in terms of its boiling and freezing points, a phenomenon known as colligative properties.
• In a solution, solvent molecules are less "free" to vaporize because they are surrounded by solute molecules.

Through understanding solute and solvent interactions, it becomes clearer why Raoult's Law states that the vapor pressure of a solution is directly proportional to the mole fraction of the solvent. The more solvent molecules there are in comparison to solute molecules, the higher the likelihood they can escape into the vapor phase.

Partial Pressure of the Solvent

Partial pressure deals with the individual pressure exerted by a substance in a mixture of gases. When we consider a solution, particularly one in equilibrium, this concept translates to the solvent's contribution to the total vapor pressure of the mixture.
Raoult's Law can be used to calculate the partial vapor pressure of the solvent. It states:\[P = P^{0} \times \left( \frac{n_{2}}{n_{1} + n_{2}} \right)\]

• \(P^{0}\) is the vapor pressure of the pure solvent.
• \(P\) is the partial vapor pressure of the solvent in the solution.
• The term \(\left( \frac{n_{2}}{n_{1} + n_{2}} \right)\) represents the mole fraction of the solvent.

Through this formula, one can determine how the presence of a solute affects the ability of the solvent to exert a gaseous pressure above the liquid. Understanding partial pressure in solutions helps clarify why certain solutions behave differently in closed systems compared to their pure solvent counterparts.