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If a is the degree of dissociation of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) the van't Hoff factor (i) used for calculating the molecular mass is (a) \(1+\alpha\) (b) \(1-\alpha\) (c) \(1+2 \alpha\) (d) \(1-2 \alpha\)

Short Answer

Expert verified
The van't Hoff factor (i) is \(1 + 2\alpha\), option (c).

Step by step solution

01

Determine the dissociation reaction

Write the dissociation reaction for \(\text{Na}_2\text{SO}_4\) in water: \[ \text{Na}_2\text{SO}_4 \rightarrow 2 \text{Na}^+ + \text{SO}_4^{2-} \].
02

Define the degree of dissociation

The degree of dissociation (\(\alpha\)) is the fraction of the original solute that dissociates into ions. If \(\alpha\) is the degree of dissociation of \(\text{Na}_2\text{SO}_4\), then \(\text{Na}_2\text{SO}_4\) dissociates into \(2\alpha\) \(\text{Na}^+\) ions and \(\alpha\) \(\text{SO}_4^{2-}\) ions.
03

Calculate the total number of particles after dissociation

Initially, we have 1 mole of \(\text{Na}_2\text{SO}_4\). After dissociation, we have \(1-\alpha\) moles of undissociated \(\text{Na}_2\text{SO}_4\), \(2\alpha\) moles of \(\text{Na}^+\), and \(\alpha\) moles of \(\text{SO}_4^{2-}\). So, the total number of moles of particles is \((1-\alpha) + 2\alpha + \alpha = 1 + 2\alpha\).
04

Define the van't Hoff factor

The van't Hoff factor (\(i\)) is the ratio of the total number of moles of particles after dissociation to the number of moles of solute before dissociation. Therefore, \(i = 1 + 2\alpha\).
05

Match with given options

Compare the calculated van't Hoff factor \(i = 1 + 2\alpha\) with the given options. The correct choice is (c) \(1 + 2\alpha\).

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Key Concepts

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

Degree of Dissociation
The degree of dissociation is an important concept in chemistry that tells us how much of a compound dissociates into its ions when dissolved in water. In simpler terms, it measures how many of the initial molecules break apart to form ions. This is represented by the symbol \( \alpha \).
It's crucial to know this because the degree of dissociation affects how a substance behaves in a solution, influencing properties like boiling point and osmotic pressure. When we say a compound like \( \text{Na}_2\text{SO}_4 \) has a degree of dissociation \( \alpha \), it implies that a fraction \( \alpha \) of all initial molecules has dissociated. This fraction then determines the number of ions produced in the solution.
Understanding the degree of dissociation helps us predict and calculate the physical and chemical properties of solutions, which is vital in various scientific and industrial applications.
Dissociation Reaction
A dissociation reaction occurs when a compound separates into its ions in a solution. It's like the compound is breaking into smaller parts. For \( \text{Na}_2\text{SO}_4 \), the dissociation reaction is organized like this:
  • \( \text{Na}_2\text{SO}_4 \rightarrow 2\text{Na}^+ + \text{SO}_4^{2-} \)
This shows that each formula unit of \( \text{Na}_2\text{SO}_4 \) gives two \( \text{Na}^+ \) ions and one \( \text{SO}_4^{2-} \) ion when it dissolves in water.
By understanding these ions, we can predict the behavior of the solution, such as how it conducts electricity or reacts with other substances. Dissociation reactions are key in chemistry because they tell us what happens to a substance once it's in solution, affecting everything from chemical reactions to laboratory analyses.
Molecular Mass Calculation
Molecular mass calculations become straightforward once the compound dissociates, as the number of dissolved ions affects the observed properties of the solution. The van't Hoff factor (\( i \)) becomes particularly significant here. It accounts for the change in particle number after dissociation.
For example, initially, there might be one mole of \( \text{Na}_2\text{SO}_4 \). However, when it dissociates, it forms \( 1 + 2\alpha \) moles of particles because of the ions formed. Each mole of particles contributes to properties like boiling point elevation or freezing point depression, which depend on the total number of particles rather than just the chemical nature of the solute.
Understanding this concept helps in accurately predicting and calculating the changes in these colligative properties, essential in both academic studies and industrial processes.
Ions in Solution
When substances like \( \text{Na}_2\text{SO}_4 \) dissolve, they release ions into the solution. This process is fundamental to understanding solutions chemistry, as ions are responsible for many properties in a solution.
In the example of \( \text{Na}_2\text{SO}_4 \), the dissociation produces \( 2\text{Na}^+ \) ions and \( \text{SO}_4^{2-} \) ions. This means for every mole of \( \text{Na}_2\text{SO}_4 \), two moles of sodium ions and one mole of sulfate ions end up in the solution.
  • Sodium ions (\( \text{Na}^+ \)) help in conducting electricity through the solution.
  • Sulfate ions (\( \text{SO}_4^{2-} \)) play roles in forming precipitates with other ions.

The behavior and interaction of these ions determine the solution’s overall characteristics and contribute to reactions and properties observable in a laboratory. Knowing how ions behave is crucial for predicting chemical behaviors, creating solutions, and analyzing results.

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Most popular questions from this chapter

An aqueous solution of \(6.3 \mathrm{~g}\) oxalic acid dihydrate is made up to \(250 \mathrm{~mL}\). The volume of \(0.1 \mathrm{~N} \mathrm{NaOH}\) required to completely neutralize \(10 \mathrm{~mL}\) of this solution is (a) \(40 \mathrm{~mL}\) (b) \(20 \mathrm{~mL}\) (c) \(10 \mathrm{~mL}\) (d) \(4 \mathrm{~mL}\)

The vapour pressure of two liquids ' \(\mathrm{P}\) ' and ' \(\mathrm{Q}\) ' are 80 and 60 torr respectively. The total vapour pressure of solution obtained by mixing 3 mole of \(\mathrm{P}\) and \(2 \mathrm{~mol}\) of Q would be (a) 20 torr (b) 72 torr (c) 68 torr (d) 140 torr

The rise in the boiling point of a solution containing \(1.8 \mathrm{~g}\) of glucose in \(100 \mathrm{~g}\) of solvent is \(0.1^{\circ} \mathrm{C}\). The molal elevation constant of the liquid is (a) \(1 \mathrm{~K} / \mathrm{m}\) (b) \(0.1 \mathrm{~K} / \mathrm{m}\) (c) \(0.01 \mathrm{~K} / \mathrm{m}\) (d) \(10 \mathrm{~K} / \mathrm{m}\)

The colligative properties of electrolytes require a slightly different approach than the one used for the colligative properties of non- electrolytes. The electrolytes dissociate into ions in solution. It is the number of solute particles that determine the colligative properties of a solution. The electrolyte solutions, therefore show abnormal colligative properties. To account for this effect we define a quantity; called the van't Hoff factor which is given by [solution] \(i=\) dissociation \(\mathrm{i}=1\) (for non - electrolytes); \(\mathrm{i}>1\) (for electrolytes, undergoing dissociation) \(\mathrm{i}<1\) (for solute, undergoing association) Benzoic acid undergoes dimerisation in benzene solution. The van't Hoff factor 'i' is related to the degree of association ' \(\alpha\) ' of the acid as (a) \(i=1+\alpha\) (b) \(i=1-\alpha\) (c) \(i=1+\frac{\alpha}{2}\) (d) \(i=1-\frac{\alpha}{2}\)

The vapour pressure of water at \(300 \mathrm{~K}\) in a closed container is \(0.4 \mathrm{~atm}\). If the volume of the container is doubled, its vapour pressure at \(300 \mathrm{~K}\) will be (a) \(0.8 \mathrm{~atm}\) (b) \(0.2 \mathrm{~atm}\) (c) \(0.4 \mathrm{~atm}\) (d) \(0.6 \mathrm{~atm}\)

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