Problem 13 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Consider separate solutions of \(0.500 \mathrm{M} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q), 0.100 \mathrm{M} \mathrm{Mg}_{3}\) \(\left(\mathrm{PO}_{4}\right)_{2}(a q), 0.250 \mathrm{M} \mathrm{KBr}(a q)\) and \(0.125 \mathrm{M}\) \(\mathrm{Na}_{3} \mathrm{PO}_{4}(a q)\) at \(25^{\circ} \mathrm{C}\). Which statement is true about these solutions, assuming all salts to be strong electrolytes? (a) They all have the same osmotic pressure. (b) \(0.100 \mathrm{M} \mathrm{Mg}_{3}\left(\mathrm{PO}_{4}\right)_{2}(a q)\) has the highest osmotic pressure. (c) \(0.125 \mathrm{M} \mathrm{Na}_{3} \mathrm{PO}_{4}(a q)\) has the highest osmotic pressure. (d) \(0.500 \mathrm{M} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q)\) has the highest osmotic pressure.

Step-by-Step Solution

Answer

(a) They all have the same osmotic pressure.

Step 1: Understanding Osmotic Pressure

The osmotic pressure of a solution is determined by the van't Hoff equation: \( \Pi = iMRT \), where \( \Pi \) is the osmotic pressure, \( i \) is the van't Hoff factor (number of particles the solute dissociates into), \( M \) is the molarity, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin.

Step 2: Evaluating Each Solution

First, we identify the van't Hoff factor \( i \) for each solution: 1. \( \text{C}_2\text{H}_5\text{OH} \) is a non-electrolyte, \( i = 1 \).2. \( \text{Mg}_3(\text{PO}_4)_2 \) dissociates into 5 ions (3 Mg\(^{2+}\) and 2 PO\(_4^{3-}\)), so \( i = 5 \).3. \( \text{KBr} \) dissociates into 2 ions, so \( i = 2 \).4. \( \text{Na}_3\text{PO}_4 \) dissociates into 4 ions (3 Na\(^+\) and 1 PO\(_4^{3-}\)), so \( i = 4 \).

Step 3: Calculating Osmotic Pressure of Each Solution

To find the osmotic pressure for each solution, calculate \( iM \) for each:1. \( 0.500 \text{ M } \text{C}_2\text{H}_5\text{OH}: iM = 1 \times 0.500 = 0.500 \)2. \( 0.100 \text{ M } \text{Mg}_3(\text{PO}_4)_2: iM = 5 \times 0.100 = 0.500 \)3. \( 0.250 \text{ M } \text{KBr}: iM = 2 \times 0.250 = 0.500 \)4. \( 0.125 \text{ M } \text{Na}_3\text{PO}_4: iM = 4 \times 0.125 = 0.500 \)

Step 4: Determining the Statement

All solutions have the calculated \( iM \) value of 0.500. Since \( RT \) is constant for each solution at the same temperature, all solutions have the same osmotic pressure. Therefore, statement (a) is true.

Key Concepts

Van't Hoff Factor

The Van't Hoff Factor, often represented by the symbol \( i \), plays a crucial role in understanding the behavior of solutions, especially when dealing with their colligative properties like osmotic pressure. It reflects the number of particles a compound dissociates into when dissolved in a solution. For non-electrolytes, such as organic compounds like ethanol \( (\text{C}_2\text{H}_5\text{OH}) \), the compound typically remains undissociated, so the Van't Hoff factor \( i \) is 1. However, for strong electrolytes, which include many ionic compounds, the Van't Hoff Factor is greater than 1. This is because these compounds dissociate into multiple ions.
For example:

• Magnesium phosphate \( (\text{Mg}_3(\text{PO}_4)_2) \) breaks down into 5 ions in solution: 3 magnesium ions \( (\text{Mg}^{2+}) \) and 2 phosphate ions \( (\text{PO}_4^{3-}) \), giving \( i = 5 \).
• Sodium phosphate \( (\text{Na}_3\text{PO}_4) \) dissociates into 4 ions: 3 sodium ions \( (\text{Na}^+) \) and 1 phosphate ion \( (\text{PO}_4^{3-}) \), resulting in \( i = 4 \).
• Potassium bromide \( (\text{KBr}) \) splits into 2 ions, a potassium ion \( (\text{K}^+) \) and a bromide ion \( (\text{Br}^-) \), leading to \( i = 2 \).

Recognizing the value of the Van't Hoff Factor is key in calculating colligative properties and understanding the solution's behavior.

Electrolyte Solutions

Electrolyte solutions are a fundamental concept in solution chemistry, significantly affecting properties like osmotic pressure. When electrolytes dissolve in water, they dissociate into charged ions, which are crucial carriers of electricity. This dissociation process increases the number of solute particles in the solution, enhancing colligative properties.
Strong electrolytes dissociate completely in solution, resulting in a high Van't Hoff Factor and therefore a higher effect on osmotic pressure and other colligative properties. For instance, salts like \( \text{Na}_3\text{PO}_4 \) and \( \text{Mg}_3(\text{PO}_4)_2 \) are categorized as strong electrolytes because of their complete dissociation into multiple ions. This results in increased osmotic pressure compared to solutions of non-electrolytes at the same concentration.
Electrolytic behavior explains why adding salt to water alters properties like boiling point, freezing point, and osmotic pressure. An important point to remember is that the type and number of ions formed during the dissociation of an electrolyte determine the magnitude of these changes.
Understanding electrolyte solutions helps predict and manipulate solution properties, important in fields like biology, chemistry, and medicine, where controlling osmotic pressure and ion concentrations is essential.

Solution Chemistry

Solution chemistry covers a broad range of topics, but at its core, it studies the properties and behaviors of solutes dissolved in solvents. When dealing with solution chemistry, the focus is often on how solutes interact with the solvent at molecular levels and how these interactions affect macroscopic properties like osmotic pressure, boiling point, and freezing point.

One key aspect of solution chemistry is the understanding of colligative properties. These properties depend largely on the number of solute particles present in the solution rather than the nature of the solute itself. This links directly to the Van't Hoff Factor, as it influences the effective concentration of particles, impacting colligative properties like osmotic pressure.

For example, if you consider a solution's osmotic pressure, calculated using the formula \( \Pi = iMRT \), where \( i \) is the Van't Hoff factor, \( M \) is molarity, \( R \) is the gas constant, and \( T \) is temperature, you can see how solution chemistry principles allow us to understand and predict the behavior of solutions.

The principles of solution chemistry not only apply to simple salt solutions but also to diverse systems involving acids, bases, and complex organic compounds in both everyday and industrial applications.