A solution is made of 2.05g of sucrose [C12H22O11] dissolved in 9.87g of water . Calculate freezing and boiling points of sucrose solution

Question

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Answer 1

To calculate the freezing and boiling points of a sucrose solution, you can use the formula for colligative properties. For the freezing point depression and boiling point elevation, you can use the following formulas:

1. **Freezing Point Depression (ΔTf):**

ΔTf = Kf * molality

2. **Boiling Point Elevation (ΔTb):**

ΔTb = Kb * molality

Where:

- ΔTf is the change in freezing point.

- ΔTb is the change in boiling point.

- Kf is the cryoscopic constant for the solvent (for water, Kf = 1.86 °C·kg/mol).

- Kb is the ebullioscopic constant for the solvent (for water, Kb = 0.512 °C·kg/mol).

- Molality is the concentration of the solute in the solution, in moles of solute per kilogram of solvent.

First, calculate the molality of the sucrose solution:

Molar mass of sucrose (C12H22O11):

= 12(12.01 g/mol) + 22(1.01 g/mol) + 11(16.00 g/mol)

= 144.12 g/mol

Number of moles of sucrose (n):

= Mass (g) / Molar mass (g/mol)

= 2.05 g / 144.12 g/mol

≈ 0.0142 moles

The mass of water doesn't change during the calculation.

Molality (m):

= Moles of solute / Mass of solvent (kg)

= 0.0142 moles / (9.87 g / 1000)

≈ 1.44 mol/kg

Now, you can calculate the freezing point depression and boiling point elevation:

1. Freezing Point Depression (ΔTf):

ΔTf = Kf * molality

ΔTf = 1.86 °C·kg/mol * 1.44 mol/kg

ΔTf ≈ 2.68 °C

So, the freezing point of the sucrose solution is lowered by approximately 2.68°C.

2. Boiling Point Elevation (ΔTb):

ΔTb = Kb * molality

ΔTb = 0.512 °C·kg/mol * 1.44 mol/kg

ΔTb ≈ 0.74 °C

Therefore, the boiling point of the sucrose solution is elevated by approximately 0.74°C.

To find the actual freezing and boiling points of the solution, you need to add/subtract these changes from the normal freezing and boiling points of pure water:

- The normal freezing point of water is 0°C.

- The normal boiling point of water is 100°C.

So, for the sucrose solution:

- Freezing point ≈ 0°C - 2.68°C ≈ -2.68°C

- Boiling point ≈ 100°C + 0.74°C ≈ 100.74°C

The freezing point of the sucrose solution is approximately -2.68°C, and the boiling point is approximately 100.74°C.