Assuming 100% dissociation, calculate the freezing point and boiling point of 2.45 m Na2SO4(aq). Constants may be found here sites. google. com/site/chempendix/colligative.

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To calculate the freezing point and boiling point of a solution, we need to use the formula for freezing point depression and boiling point elevation, which are colligative properties that depend on the molality of the solution.

Let's break down the steps to find the freezing and boiling points:

Step 1: Calculate the molality of the solution.
Molality (m) is defined as the number of moles of solute per kilogram of solvent. In this case, the solute is Na2SO4 and the solvent is water.

Given:
Mass of Na2SO4 = 2.45 m
Molar mass of Na2SO4 = 22.99 g/mol + (32.07 g/mol * 2) + (16.00 g/mol * 4) = 142.04 g/mol
Density of water = 1.00 g/mL = 1.00 kg/L

First, calculate the number of moles of Na2SO4:
Moles of Na2SO4 = (mass of Na2SO4) / (molar mass of Na2SO4)
Moles of Na2SO4 = (2.45 mol) / (142.04 g/mol) = 0.017 mol

Next, calculate the mass of water:
Mass of water = (mass of solution) - (mass of Na2SO4)
Mass of water = (2.45 kg) - (0.017 kg) = 2.433 kg

Finally, calculate the molality:
Molality = (moles of Na2SO4) / (mass of water in kg)
Molality = (0.017 mol) / (2.433 kg) = 0.00698 m

Step 2: Use the colligative property equations to find the freezing and boiling points.

Freezing point depression formula:
ΔTf = Kf * m
where ΔTf is the change in freezing point, Kf is the molal freezing point depression constant, and m is the molality of the solution.

Boiling point elevation formula:
ΔTb = Kb * m
where ΔTb is the change in boiling point, Kb is the molal boiling point elevation constant, and m is the molality of the solution.

Given constants:
For Na2SO4 in water, Kf = 1.86 °C/m and Kb = 0.512 °C/m (taken from the provided website).

Now, let's calculate the freezing point depression and boiling point elevation:

ΔTf = Kf * m
ΔTf = 1.86 °C/m * 0.00698 m = 0.01299 °C

ΔTb = Kb * m
ΔTb = 0.512 °C/m * 0.00698 m = 0.00357 °C

Step 3: Calculate the freezing and boiling points.

The freezing point of a solution is lower than the freezing point of the pure solvent, while the boiling point is higher than the boiling point of the pure solvent.

For water, the freezing point is 0 °C and the boiling point is 100 °C.

Freezing point = 0 °C - ΔTf = 0 °C - 0.01299 °C = -0.01299 °C (rounded to -0.01 °C)

Boiling point = 100 °C + ΔTb = 100 °C + 0.00357 °C = 100.00357 °C (rounded to 100.00 °C)

Therefore, the freezing point of a 2.45 m Na2SO4(aq) solution assuming 100% dissociation is approximately -0.01 °C, and the boiling point is approximately 100.00 °C.