Read the Passage and answer the following question:

There are many properties of ideal solutions which depends only on the concentration of the solute particles and are independent of the nature of the solute. Such properties are called colligative properties. For example, when a non-volatile solute is added to a solvent, its vapour pressure gets lowered. Raoult established that relative lowering in vapour pressure depends only on the concentration of the solute and it is independent of its identity. Similarly, elevation in boiling point, depression in freezing point and osmotic pressure on addition of a non volatile solute to a solvent depend only on the concentration of solute but are independent of its nature. These properties can also be used to determine the molar mass of the solute. The relative lowering in vapour pressure changes when the solute is electrolyte and undergoes association or dissociation in the solution. The molar mass obtained in these electrolytic solutions are abnormal molar masses. 

What will be the relation between osmotic pressure at 273 K if 12 g of urea (p₁), 12 g of glucose (p₂) and 12 g of sucrose (p₃) are dissolved in the same amount of water?

p₁>p₂>p₃

The correct answer is option 2. p₁>p₂>p₃.

Osmotic pressure (\(\Pi\)) of a solution is given by the formula:

\(\Pi = iMRT \)

where:

\(i\) is the van 't Hoff factor (which is 1 for non-electrolytes like urea, glucose, and sucrose),

\(M\) is the molarity of the solution,

\(R\) is the gas constant,

\(T\) is the temperature in Kelvin.

Since urea, glucose, and sucrose do not dissociate in water, their van 't Hoff factors (\(i\)) are all 1. Therefore, the osmotic pressure depends directly on the molarity (\(M\)) of the solute.

Molarity (\(M\)) is calculated as:

\(M = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \)

Since the volume of the solution is the same for each solute, the molarity will be determined by the number of moles of solute.

To find the number of moles, we use the formula:

\(\text{moles} = \frac{\text{mass}}{\text{molar mass}} \)

Given:

Mass of urea = 12 g, molar mass of urea (\(\text{CO(NH}_2\text{)}_2\)) = 60 g/mol,

Mass of glucose = 12 g, molar mass of glucose (\(\text{C}_6\text{H}_{12}\text{O}_6\)) = 180 g/mol,

Mass of sucrose = 12 g, molar mass of sucrose (\(\text{C}_{12}\text{H}_{22}\text{O}_{11}\)) = 342 g/mol.

Moles of urea: \(\frac{12 \text{ g}}{60 \text{ g/mol}} = 0.2 \text{ mol}\)

Moles of glucose: \(\frac{12 \text{ g}}{180 \text{ g/mol}} = 0.0667 \text{ mol}\)

Moles of sucrose: \(\frac{12 \text{ g}}{342 \text{ g/mol}} = 0.0351 \text{ mol}\)

Since osmotic pressure is directly proportional to the number of moles of solute, we can compare the osmotic pressures:

Osmotic pressure of urea (\(p1\)) will be the highest since it has the highest number of moles (0.2 mol),

Osmotic pressure of glucose (\(p2\)) will be next since it has fewer moles than urea but more than sucrose (0.0667 mol),

Osmotic pressure of sucrose (\(p3\)) will be the lowest since it has the least number of moles (0.0351 mol).

Therefore, the relation between the osmotic pressures is 2. p1 > p2 > p3.