Calculate the boiling point elevation of a solution containing 22.8 g of glucose (C6H12O6) dissolved in 672.0 g of water. Calculate the freezing point depression for the same solution.

To calculate the boiling point elevation and freezing point depression of a solution, we need to use the formulas provided by colligative properties.

1. Boiling Point Elevation:
The boiling point elevation (∆Tb) is given by the formula:

∆Tb = Kb * m

where ∆Tb is the boiling point elevation, Kb is the molal boiling point constant, and m is the molality of the solute.

In this case, the solute is glucose (C6H12O6) and the solvent is water (H2O). We first need to calculate the molality (m) of the solution, which is defined as the number of moles of solute per kilogram of solvent.

Step 1: Calculate the number of moles of glucose (C6H12O6):
mol = mass / molar mass
mol = 22.8 g / 180.18 g/mol (molar mass of glucose)
mol ≈ 0.127 mol

Step 2: Calculate the molality of the solution:
molality (m) = mol solute / kg solvent
kg solvent = 672.0 g / 1000 g/kg (since the mass is given in grams)
kg solvent = 0.672 kg

m = 0.127 mol / 0.672 kg
m ≈ 0.189 mol/kg

Step 3: Calculate the boiling point elevation (∆Tb):
Given that the molal boiling point constant (Kb) for water is approximately 0.512 °C/m, we can calculate the boiling point elevation:

∆Tb = 0.512 °C/m * 0.189 mol/kg
∆Tb ≈ 0.097 °C

Therefore, the boiling point of the solution will be elevated by approximately 0.097 °C.

2. Freezing Point Depression:
The freezing point depression (∆Tf) is given by the formula:

∆Tf = Kf * m

where ∆Tf is the freezing point depression, Kf is the molal freezing point constant, and m is the molality of the solute.

For water, the molal freezing point constant (Kf) is approximately 1.86 °C/m. We can use the same molality (m) calculated earlier.

∆Tf = 1.86 °C/m * 0.189 mol/kg
∆Tf ≈ 0.352 °C

Therefore, the freezing point of the solution will be depressed by approximately 0.352 °C.

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