Problem 41 - A Complete Resource Book in Chemistry for JEE Main

Question

The relationship between osmotic pressure at 273 \(\mathrm{K}\) when \(10 \mathrm{~g}\) glucose \(\left(\mathrm{P}_{1}\right) 10 \mathrm{~g}\) urea \(\left(\mathrm{P}_{2}\right)\) and \(10 \mathrm{~g}\) sucrose \(\left(\mathrm{P}_{3}\right)\) are dissolved in \(250 \mathrm{~mL}\) of water is (a) \(\mathrm{P}_{1}>\mathrm{P}_{2}>\mathrm{P}_{3}\) (b) \(\mathrm{P}_{3}>\mathrm{P}_{2}>\mathrm{P}_{1}\) (c) \(\mathrm{P}_{2}>\mathrm{P}_{1}>\mathrm{P}_{3}\) (d) \(\mathrm{P}_{2}>\mathrm{P}_{3}>\mathrm{P}_{1}\)

Step-by-Step Solution

Answer

Option (c) \( \mathrm{P}_{2} > \mathrm{P}_{1} > \mathrm{P}_{3} \).

Step 1: Determine Molar Masses

First, let's determine the molar masses of each solute:- **Glucose (\(C_6H_{12}O_6\))**: \( (6 \times 12.01) + (12 \times 1.008) + (6 \times 16.00) = 180.18 \, \text{g/mol} \)- **Urea (\(CH_4N_2O\))**: \( (1 \times 12.01) + (4 \times 1.008) + (2 \times 14.01) + (1 \times 16.00) = 60.06 \, \text{g/mol} \)- **Sucrose (\(C_{12}H_{22}O_{11}\))**: \( (12 \times 12.01) + (22 \times 1.008) + (11 \times 16.00) = 342.30 \, \text{g/mol} \)

Step 2: Calculate Moles for Each Solute

Use the molar masses found in Step 1 to calculate the moles of each solute:- **Moles of Glucose**: \( \frac{10}{180.18} = 0.0555 \, \text{mol} \)- **Moles of Urea**: \( \frac{10}{60.06} = 0.1665 \, \text{mol} \)- **Moles of Sucrose**: \( \frac{10}{342.30} = 0.0292 \, \text{mol} \)

Step 3: Apply Concept of Osmotic Pressure

Osmotic pressure (\(\Pi\)) is calculated using the formula: \( \Pi = i \cdot n \cdot R \cdot T \), where \(i\) is the van't Hoff factor (1 for non-electrolytes like glucose, urea, and sucrose), \(n\) is the moles, \(R\) is the gas constant, and \(T\) is the temperature.Since all are non-electrolytes, consider only moles: \(\Pi \propto n\), which means osmotic pressure is directly proportional to the number of moles.

Step 4: Compare Osmotic Pressures

With osmotic pressure proportional to moles, we rank them:- **Glucose**: \(0.0555 \, \text{mol} \)- **Urea**: \(0.1665 \, \text{mol} \)- **Sucrose**: \(0.0292 \, \text{mol} \)Hence, osmotic pressures are ranked: \(\mathrm{P}_2 > \mathrm{P}_1 > \mathrm{P}_3\).

Key Concepts

Understanding Molar Mass

Molar mass is an essential concept in chemistry that determines the mass of one mole of a given substance. A mole is a standard unit used to measure chemical substances. The molar mass is usually expressed in grams per mole (g/mol) and provides the mass of a substance for a specific amount, in this case, one mole.

To calculate the molar mass, you sum up the atomic masses of all atoms in a molecule. For example:

• **Glucose (\(C_6H_{12}O_6\)**): The molar mass is calculated as \((6 \times 12.01) + (12 \times 1.008) + (6 \times 16.00) = 180.18 \, \text{g/mol}\).
• **Urea (\(CH_4N_2O\)**): Here, the molar mass is \((1 \times 12.01) + (4 \times 1.008) + (2 \times 14.01) + (1 \times 16.00) = 60.06 \, \text{g/mol}\).
• **Sucrose (\(C_{12}H_{22}O_{11}\)**): For sucrose, it comes out to be \((12 \times 12.01) + (22 \times 1.008) + (11 \times 16.00) = 342.30 \, \text{g/mol}\).

Molar mass is a crucial step because it helps us determine how much of a substance is present in a given mass, which is important when calculating other properties such as moles and osmotic pressure.

The Role of van't Hoff Factor

The van't Hoff factor, denoted as \(i\), is a crucial parameter used in calculating osmotic pressure. It indicates the number of particles a solute splits into or forms when it is dissolved in a solvent. For electrolytes, which dissociate into ions, \(i\) can be greater than 1. For non-electrolytes, which do not dissociate, the van't Hoff factor is typically 1.

In our exercise, the substances glucose, urea, and sucrose are non-electrolytes. This means that they do not dissociate into ions in the solution, keeping their molecular form intact. Thus, their van't Hoff factor is effectively 1.
Understanding the van't Hoff factor for non-electrolytes is important because it simplifies the formula for osmotic pressure, leading to \(\Pi = i \cdot n \cdot R \cdot T = n \cdot R \cdot T\). This allows us to focus solely on the moles of solute when considering the osmotic pressure.

Recognizing Non-electrolytes

Non-electrolytes are substances that do not dissociate into ions when dissolved in water. This characteristic means they do not conduct electricity in solution, unlike electrolytes, which do break into ions. Identifying non-electrolytes, like glucose, urea, and sucrose in this exercise, allows us to treat them differently in calculations such as osmotic pressure.

Non-electrolytes dissolve in water without changing their molecular structure. They simply disperse as intact molecules in the solution. Because of this property, their contribution to the osmotic pressure is straightforward—they exert osmotic pressure proportional to their number of moles in the solution.
When dealing with colligative properties like osmotic pressure, understanding non-electrolytes is helpful because it means you don't need to adjust their effect on the property beyond simply counting their presence as whole molecules.

Proportionality of Osmotic Pressure to Moles

In the realm of chemistry, osmotic pressure (\(\Pi\)) is a colligative property that depends on the number of particles in a solution, not their identity. For non-electrolytes, where the van't Hoff factor is 1, osmotic pressure is directly proportional to the number of moles of solute present. This is expressed in the relation \(\Pi = n \cdot R \cdot T\), where:

• \(n\) is the number of moles of solute,
• \(R\) is the universal gas constant, and
• \(T\) is the temperature in Kelvin.

The exercise shows how **glucose**, **urea**, and **sucrose** dissolve and create solutions each with unique osmotic pressures. This is due mainly to the different amounts of moles formed by dissolving 10 grams of each solute, as calculated. The result demonstrates that urea has the highest osmotic pressure, followed by glucose, and then sucrose, due purely to the differences in their moles.
Recognizing that osmotic pressure is not dependent on the type of solute, but rather on the number of particles, provides a clearer understanding of solution behavior and is pivotal when analyzing experiments and real-world applications.