calculate the freezing point of following aqueous solution.

1. A solution containing 10g of urea in 500g of water.

2. A solution containing 0.03kg of sucrose in 750 g of water.

3. A solution containing 8 g of glucose in 0.3 kg of water ( use kf of water = 1.86).

To calculate the freezing point of the given aqueous solutions, we can use the formula for freezing point depression, which is expressed as:

ΔTf = i * Kf * m

Where:

• ΔTf = change in freezing point
• i = van 't Hoff factor (number of particles the solute breaks into)
• Kf = freezing point depression constant of the solvent (for water, Kf = 1.86 °C kg/mol)
• m = molality of the solution (moles of solute per kg of solvent)

Let’s break down the calculations for each solution step by step.

1. Urea Solution

For urea, the van 't Hoff factor (i) is 1 because it does not dissociate in solution.

• Mass of urea: 10 g
• Molar mass of urea (NH2CONH2): approximately 60 g/mol
• Mass of water: 500 g = 0.5 kg

First, calculate the number of moles of urea:

moles of urea = mass / molar mass = 10 g / 60 g/mol = 0.1667 mol

Next, calculate the molality (m):

m = moles of solute / kg of solvent = 0.1667 mol / 0.5 kg = 0.3334 mol/kg

Now, apply the freezing point depression formula:

ΔTf = i * Kf * m = 1 * 1.86 °C kg/mol * 0.3334 mol/kg = 0.620 °C

The freezing point of pure water is 0 °C, so:

Freezing point of the solution = 0 °C - 0.620 °C = -0.620 °C

2. Sucrose Solution

Sucrose also has a van 't Hoff factor of 1 since it does not dissociate.

• Mass of sucrose: 0.03 kg = 30 g
• Molar mass of sucrose (C12H22O11): approximately 342 g/mol
• Mass of water: 750 g = 0.75 kg

Calculate the number of moles of sucrose:

moles of sucrose = 30 g / 342 g/mol = 0.0877 mol

Now, find the molality:

m = 0.0877 mol / 0.75 kg = 0.1169 mol/kg

Using the freezing point depression formula:

ΔTf = 1 * 1.86 °C kg/mol * 0.1169 mol/kg = 0.217 °C

Thus, the freezing point of the solution is:

Freezing point = 0 °C - 0.217 °C = -0.217 °C

3. Glucose Solution

Glucose also has a van 't Hoff factor of 1.

• Mass of glucose: 8 g
• Molar mass of glucose (C6H12O6): approximately 180 g/mol
• Mass of water: 0.3 kg

Calculate the number of moles of glucose:

moles of glucose = 8 g / 180 g/mol = 0.0444 mol

Next, find the molality:

m = 0.0444 mol / 0.3 kg = 0.148 mol/kg

Now, apply the freezing point depression formula:

ΔTf = 1 * 1.86 °C kg/mol * 0.148 mol/kg = 0.275 °C

The freezing point of the glucose solution is:

Freezing point = 0 °C - 0.275 °C = -0.275 °C

Summary of Freezing Points

• Urea solution: -0.620 °C
• Sucrose solution: -0.217 °C
• Glucose solution: -0.275 °C

These calculations illustrate how the addition of solutes like urea, sucrose, and glucose affects the freezing point of water, demonstrating the concept of freezing point depression in solutions.