A solution of sucrose (Molar mass 342 g mol-1) is prepared by dissolving 68.4 g in 1000 g of water. What is the (i) Vapour pressure of the solution at 293k . (ii) Osmotic pressure at 293k. (iii) Boiling point of the solution. (iv) Freezing point of the solution.

To tackle your question about the properties of a sucrose solution, we need to apply some fundamental concepts from colligative properties, which are properties that depend on the number of solute particles in a solution rather than their identity. Let's break this down step by step for each part of your question.

Calculating the Vapor Pressure of the Solution

The vapor pressure of a solution can be determined using Raoult's Law, which states that the vapor pressure of a solvent in a solution is equal to the vapor pressure of the pure solvent multiplied by its mole fraction in the solution.

Step 1: Calculate Moles of Sucrose

First, we need to find the number of moles of sucrose:

• Molar mass of sucrose = 342 g/mol
• Moles of sucrose = mass / molar mass = 68.4 g / 342 g/mol = 0.2 moles

Step 2: Calculate Moles of Water

Next, we calculate the moles of water:

• Molar mass of water = 18 g/mol
• Moles of water = 1000 g / 18 g/mol = 55.56 moles

Step 3: Calculate Mole Fraction of Water

Now, we can find the mole fraction of water:

• Total moles = moles of sucrose + moles of water = 0.2 + 55.56 = 55.76 moles
• Mole fraction of water = moles of water / total moles = 55.56 / 55.76 ≈ 0.995

Step 4: Apply Raoult's Law

If we assume the vapor pressure of pure water at 293 K is approximately 23.8 mmHg, we can calculate the vapor pressure of the solution:

• Vapor pressure of solution = vapor pressure of pure water × mole fraction of water
• Vapor pressure of solution = 23.8 mmHg × 0.995 ≈ 23.65 mmHg

Osmotic Pressure Calculation

The osmotic pressure can be calculated using the formula:

• Osmotic pressure (π) = iCRT

For sucrose, which does not dissociate in solution, i = 1. The values are:

• C = moles of solute / volume of solution in liters. Assuming the volume is approximately 1 L (since the mass of water is 1000 g), C = 0.2 moles / 1 L = 0.2 mol/L.
• R = 0.0821 L·atm/(K·mol)
• T = 293 K

Now, substituting these values:

• π = 1 × 0.2 mol/L × 0.0821 L·atm/(K·mol) × 293 K ≈ 4.87 atm

Boiling Point Elevation

The boiling point elevation can be calculated using the formula:

• ΔT_b = iK_bm

Where:

• K_b for water = 0.512 °C kg/mol
• m = molality = moles of solute / kg of solvent = 0.2 moles / 1 kg = 0.2 mol/kg

Now, substituting the values:

• ΔT_b = 1 × 0.512 °C kg/mol × 0.2 mol/kg = 0.1024 °C
• Boiling point of solution = 100 °C + 0.1024 °C ≈ 100.1 °C

Freezing Point Depression

For freezing point depression, we use the formula:

• ΔT_f = iK_fm

Where:

• K_f for water = 1.86 °C kg/mol

Substituting the values:

• ΔT_f = 1 × 1.86 °C kg/mol × 0.2 mol/kg = 0.372 °C
• Freezing point of solution = 0 °C - 0.372 °C ≈ -0.372 °C

Summary of Results

• Vapor Pressure: Approximately 23.65 mmHg
• Osmotic Pressure: Approximately 4.87 atm
• Boiling Point: Approximately 100.1 °C
• Freezing Point: Approximately -0.372 °C

These calculations illustrate how colligative properties can significantly alter the physical characteristics of a solution. If you have any further questions or need clarification on any of these steps, feel free to ask!