Which one of the following solution has least vapour pressure? (a) \(0.01 \mathrm{M} \mathrm{CaCl}_{2}\) (b) \(0.01 \mathrm{M}\) glucose (c) \(0.01 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\) (d) \(0.01 \mathrm{M} \mathrm{Na}_{3} \mathrm{PO}_{4}\)

Short Answer

Expert verified
0.01 M Na3PO4 has the least vapour pressure due to the highest van't Hoff factor.

Step by step solution

01

Understand the concept of vapor pressure

Vapor pressure is the pressure exerted by the vapor in equilibrium with its liquid or solid phase. For solutions, the vapor pressure is influenced by the presence of solute particles because they can reduce the vapor pressure of the solvent. This is known as Raoult's Law.
02

Apply Raoult's Law

According to Raoult's Law, the vapor pressure of a solution is directly proportional to the mole fraction of the solvent. Adding solute particles decreases the mole fraction of the solvent, thus decreasing the vapor pressure of the solution. This effect is more pronounced with the more solute particles present.
03

Calculate the van't Hoff factor

The van't Hoff factor, i, represents the number of particles a compound dissociates into in solution. Glucose doesn't dissociate, so its i is 1. For ionic compounds: CaCl2 (i = 3), Na2SO4 (i = 3), and Na3PO4 (i = 4).
04

Determine which solution has the least vapor pressure

Since the vapor pressure decreases with increasing solute particles, the solution with the highest van't Hoff factor will have the least vapor pressure. In this case, Na3PO4, with the highest i value of 4, will dissociate into the most particles and therefore have the least vapor pressure.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Raoult's Law
Understanding Raoult's Law is crucial when studying solutions' vapor pressure. It asserts that the vapor pressure of the solvent in an ideal solution is directly proportional to the mole fraction of the solvent. In mathematical terms, the law is expressed as:
\( P = P^\ast \times X \)
where \( P \) is the vapor pressure of the solution, \( P^\ast \) is the vapor pressure of the pure solvent, and \( X \) is the mole fraction of the solvent. In simple terms, as the amount of solute in a solution increases, the mole fraction of the solvent decreases, leading to a decrease in the solution's vapor pressure. This is because the solute molecules occupy space at the surface of the solution, preventing as many solvent molecules from evaporating.

This intimate connection between vapor pressure and mole fraction illustrates the importance of Raoult's Law in predicting how solutions will behave under different conditions, such as during the process of distillation or in chemical equilibrium scenarios.
van't Hoff factor
The van't Hoff factor, denoted as \( i \), is a dimensionless quantity used in chemistry to account for the effect of solute particles on various colligative properties, such as boiling point elevation, freezing point depression, osmotic pressure, and, pertinent to this exercise, vapor pressure. The van't Hoff factor essentially tells us how many moles of particles (ions or molecules) a substance will produce in solution.

For example, a non-electrolyte like glucose doesn't dissociate in solution, so its \( i \) is 1. On the other hand, electrolytes like \( CaCl_2 \) dissociate into three ions (\( Ca^{2+} \) and two \( Cl^- \) ions), so its \( i \) is 3. Consequently, the presence of more particles from dissociation leads to a greater reduction in vapor pressure. Understanding the van't Hoff factor is vital for predicting a solution’s behavior, particularly in scenarios where the dissociation of compounds significantly influences the solution’s properties.
Mole fraction
The mole fraction is a way of expressing the concentration of a component in a mixture. Defined as the ratio of the moles of a specific component to the total moles of all components in the mixture, the mole fraction is a unitless quantity that provides insight into the composition of solutions. Given the symbol \( X \), it is calculated for a substance A in a solution as:
\( X_A = \frac{n_A}{n_{total}} \)
where \( n_A \) is the number of moles of the substance A, and \( n_{total} \) is the sum of the moles of all the substances in the mixture. The mole fraction does not change with temperature or pressure, distinguishing it from other concentration measures like molarity.

In the context of vapor pressure, the mole fraction effectively determines the extent to which the solvent’s vapor pressure will be lowered upon the addition of a solute. As the solute is added, the mole fraction of the solvent decreases, which, as Raoult’s Law describes, results in a lower vapor pressure for the solution. This is a key concept to grasp when identifying the volatility of components in a solution and predicting how mixtures will behave in different thermodynamic processes.

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Most popular questions from this chapter

The elevation in boiling point for \(13.44 \mathrm{~g}\) of \(\mathrm{CuCl}_{2}\) dissolved in \(1 \mathrm{~kg}\) of water as solvent will be \(\left(\mathrm{K}_{\mathrm{b}}=0.52 \mathrm{~kg} / \mathrm{J}\right.\), molar mass of \(\mathrm{CuCl}_{2}=134.4 \mathrm{~g} / \mathrm{mol}\) ) (a) \(0.05\) (b) \(0.10\) (c) \(0.16\) (d) \(0.20\).

The molality of 1 litre solution of \(93 \% \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{w} / \mathrm{v})\) having density \(1.84 \mathrm{~g} / \mathrm{mL}\) is: (a) \(10.43\) (b) \(1.043\) (c) \(0.1043\) (d) \(100.43\)

In a mixture of \(\mathrm{A}\) and \(\mathrm{B}\), components show negative deviation when: (a) \(\mathrm{A}-\mathrm{B}\) interaction is stronger than \(\mathrm{A}-\mathrm{A}\) and \(\mathrm{B}-\mathrm{B}\) interaction (b) \(\mathrm{A}-\mathrm{B}\) interaction is weaker than \(\mathrm{A}-\mathrm{A}\) and \(\mathrm{B}-\mathrm{B}\) interaction (c) \(\Delta \mathrm{V}_{\operatorname{mix}}>0, \Delta \mathrm{S}_{\operatorname{mix}}>0\) (d) \(\Delta \mathrm{V}_{\operatorname{mix}}^{\operatorname{mix}}=0, \Delta S_{\operatorname{mix}}^{\operatorname{lix}}>0\)

A decimolar solution of potassium ferrocyanide is \(50 \%\) dissociated at \(300 \mathrm{~K}\). Calculate the osmotic pressure of the solution. \(\left(\mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right)\) (a) \(0.07389 \mathrm{~atm}\) (b) \(7.389 \mathrm{~atm}\) (c) \(738.89 \mathrm{~atm}\) (d) \(73.89 \mathrm{~atm}\)

When mercuric iodide is added to the aqueous solution of potassium iodide, the: (a) Freezing point is raised (b) Freezing point does not change (c) Freezing point is lowered (d) Boiling point does not change

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