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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

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(c) \(1 + 2\alpha\)

Step by step solution

01

Understanding the concept

When a compound undergoes dissociation in a solution, it breaks down into its constituent ions. The van't Hoff factor \(i\) is used to correct the expected colligative properties for this dissociation. It represents the number of particles the compound dissociates into when dissolved.
02

Writing the dissociation equation

Sodium sulfate, \(\mathrm{Na}_{2} \mathrm{SO}_{4}\), dissociates in solution as follows: \[\mathrm{Na}_{2} \mathrm{SO}_{4} \rightarrow 2 \mathrm{Na}^{+} + \mathrm{SO}_{4}^{2-}\]. It dissociates into 3 particles total: 2 sodium ions and 1 sulfate ion.
03

Define degree of dissociation

The degree of dissociation, \(\alpha\), tells us the fraction of the original solute that has dissociated. If \(\alpha\) is the degree of dissociation, it means that \(\alpha\) portion of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) has dissociated into ions.
04

Calculate the van't Hoff factor

For a solute like \(\mathrm{Na}_{2} \mathrm{SO}_{4}\), dissociating into 3 ions, the van't Hoff factor \(i\) can be calculated as: \[ i = 1 + (n-1)\alpha \] where \(n\) is the total number of particles formed from one molecule of solute. Here, \(n = 3\), so the expression becomes: \[ i = 1 + (3-1)\alpha = 1 + 2\alpha \]

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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, denoted as \(\alpha\), is a measure of how much a compound dissociates into ions in a solution. Let's think of it as a way to express what fraction of the compound has broken up. When you dissolve a certain amount of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\), and \(\alpha\) represents the portion that dissociates, it means that \(\alpha \times 100\%\) of the sodium sulfate breaks down into ions. This helps us understand the behavior of the solution since dissociation impacts properties like boiling point and freezing point. It's essential to know \(\alpha\) to predict changes in these properties.
Colligative Properties
Colligative properties are unique because they depend on the number of particles in a solution, not the type of particles. Examples include changes in boiling and freezing points, vapor pressure lowering, and osmotic pressure.
When a solute like sodium sulfate dissociates, it affects these properties because it increases the number of particles in the solution. So, if a compound dissociates into more ions, it will have a more significant effect on colligative properties. Understanding these properties is vital for applications in chemistry, such as calculating the van't Hoff factor and consequently the molecular mass of the solute in a solution.
Molecular Mass Calculation
Calculating molecular mass using colligative properties involves using van't Hoff factor, \(i\), which adjusts for dissociation. When a solute dissociates in a solution, the apparent number of particles increases, modifying the solution's colligative properties.
For example, with sodium sulfate dissociating into three ions \([2 \mathrm{Na}^{+} + \mathrm{SO}_{4}^{2-}]\), the van't Hoff factor is calculated as \(i = 1 + 2\alpha\). This factor is then used in formulas to determine the actual molecular mass of the solute. Accurate molecular mass calculations require considering dissociation, making \(i\) a critical part of the process.
Sodium Sulfate Dissociation
Sodium sulfate, \(\mathrm{Na}_{2} \mathrm{SO}_{4}\), is comprised of sodium and sulfate ions. When placed in water, it fully dissociates into 2 sodium ions \((\mathrm{Na}^{+})\) and 1 sulfate ion \((\mathrm{SO}_{4}^{2-})\). This results in a total of three particles, affecting the solution's properties significantly.
Understanding this dissociation is crucial in many chemistry calculations. For example, it helps accurately determine how solutions will behave in terms of colligative properties and aids in computing the van't Hoff factor for these purposes. By being aware of how sodium sulfate dissociates, students can make more accurate predictions and calculations in their study of chemistry.

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Most popular questions from this chapter

On mixing, heptane and octane form an ideal solution. At \(373 \mathrm{~K}\), the vapour pressures of the two liquid components (heptane and octane) are \(105 \mathrm{kPa}\) and \(45 \mathrm{kPa}\) respectively. Vapour pressure of the solution obtained by mixing \(25.0 \mathrm{~g}\) of heptane and \(35 \mathrm{~g}\) of octane will be (molar mass of heptane \(=100 \mathrm{~g} \mathrm{~mol}^{-1}\) and of octane \(=\) \(114 \mathrm{~g} \mathrm{~mol}^{-1}\) ) (a) \(72.0 \mathrm{kPa}\) (b) \(36.1 \mathrm{kPa}\) (c) \(96.2 \mathrm{kPa}\) (d) \(144.5 \mathrm{kPa}\)

What is the freezing point of one litre of an aqueous solution of a non- electrolyte having an osmotic pressure of \(2 \mathrm{~atm}\) at \(300 \mathrm{~K}\). (a) \(273.349 \mathrm{~K}\) (b) \(263.159 \mathrm{~K}\) (c) \(215.569 \mathrm{~K}\) (d) \(272.849 \mathrm{~K}\)

Which pair of the following will not form an ideal solution? (a) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}\) (b) \(\mathrm{H}_{2} \mathrm{O}+\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{OH}\) (c) \(\mathrm{CCl}_{4}+\mathrm{SiCl}_{4}\) (d) \(\mathrm{C}_{6} \mathrm{H}_{14}+\mathrm{C}_{7} \mathrm{H}_{16}\)

During osmosis, flow of water through a semipermeable membrane is (a) from both sides of semi-permeable membrane with unequal flow rates (b) from solution having lower concentration only (c) from solution having higher concentration only (d) from both sides of semi-permeable membrane with equal flow rates

\(18 \mathrm{~g}\) of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) is added to \(178.2 \mathrm{~g}\) of water. The vapour pressure of water for this aqueous solution at \(100^{\circ} \mathrm{C}\) is (a) \(759.00\) torr (b) \(7.60\) torr (c) \(76.00\) torr (d) \(752.40\) torr

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