What is the freezing point of each solution?
a)1.40mol Na2SO4 in 1750 g H2O
b)0.060 mol MgSO4 in 100g H2O
molality = moles/kg solvent
Solve for molality.
delta T = i*Kf*m
Solve for delta T and subtract from the normal freezing point.
i is the van't Hoff factor and is the number of dissolved particles. For example, for MgSO4 that will be two since MgSO4 dissociates into Mg^+2 and SO4^=.
To determine the freezing point of each solution, we need to calculate the molality of the solute in each case.
The formula for molality (m) is:
molality (m) = moles of solute / mass of solvent in kg
a) For the first solution, we have 1.40 mol of Na2SO4 in 1750 g of H2O.
To convert the mass of water to kg, we divide by 1000: 1750 g H2O = 1.75 kg H2O.
Now we can calculate the molality:
molality (m) = 1.40 mol Na2SO4 / 1.75 kg H2O ≈ 0.8 m
b) For the second solution, we have 0.060 mol of MgSO4 in 100 g of H2O.
Again, we need to convert the mass of water to kg: 100 g H2O = 0.1 kg H2O.
Now we can calculate the molality:
molality (m) = 0.060 mol MgSO4 / 0.1 kg H2O = 0.6 m
Once we have the molality of each solution, we can use the formula for the change in freezing point:
ΔTf = Kf * m
Where:
ΔTf = change in freezing point
Kf = molal freezing point depression constant (constant for the solvent)
m = molality of the solute
For pure water, Kf = 1.86 °C/m.
a) For the first solution, with a molality of 0.8 m:
ΔTf = 1.86 °C/m * 0.8 m ≈ 1.49 °C
b) For the second solution, with a molality of 0.6 m:
ΔTf = 1.86 °C/m * 0.6 m ≈ 1.12 °C
To find the freezing point of each solution, we subtract the temperature change (ΔTf) from the freezing point of pure water (0 °C).
a) Freezing point of solution a = 0 °C - 1.49 °C ≈ -1.49 °C
b) Freezing point of solution b = 0 °C - 1.12 °C ≈ -1.12 °C
Therefore, the freezing point of solution (a) is approximately -1.49 °C, and the freezing point of solution (b) is approximately -1.12 °C.
To determine the freezing point of a solution, we need to use the concept of freezing point depression. The freezing point depression is a colligative property, which means it depends on the number of solute particles in the solution, rather than the nature of the solute itself.
The formula to calculate the freezing point depression is:
ΔTf = Kf · m
Where:
ΔTf is the freezing point depression (the difference between the freezing point of the solvent and the freezing point of the solution),
Kf is the cryoscopic constant (a constant value characteristic of the solvent),
m is the molality of the solution (the number of moles of solute per kilogram of solvent).
By rearranging the formula, we can solve for the freezing point (Tf):
Tf = Tf₀ - ΔTf
Where Tf₀ is the freezing point of the pure solvent.
Now, let's calculate the freezing point for the given solutions:
a) 1.40 mol Na2SO4 in 1750 g H2O:
First, we need to find the molality (m):
Molality (m) = moles of solute / mass of solvent (in kg)
We convert the mass of water from grams to kilograms by dividing by 1000:
m = 1.40 mol / (1750 g / 1000) kg
m = 0.80 mol/kg
Next, we determine the cryoscopic constant for water (Kf). For water, Kf is approximately 1.86 °C/m.
ΔTf = Kf · m
ΔTf = 1.86 °C/m * 0.80 mol/kg
ΔTf = 1.49 °C
Assuming the freezing point of pure water is 0 °C, we can find the freezing point of the solution:
Tf = Tf₀ - ΔTf
Tf = 0 °C - 1.49 °C
Tf ≈ -1.49 °C
Therefore, the freezing point of the solution with 1.40 mol Na2SO4 in 1750 g H2O is approximately -1.49 °C.
b) 0.060 mol MgSO4 in 100 g H2O:
First, we calculate the molality (m):
m = 0.060 mol / (100 g / 1000) kg
m = 0.60 mol/kg
Next, we determine the cryoscopic constant for water (Kf):
ΔTf = Kf · m
ΔTf = 1.86 °C/m * 0.60 mol/kg
ΔTf = 1.12 °C
Assuming the freezing point of pure water is 0 °C, we can find the freezing point of the solution:
Tf = Tf₀ - ΔTf
Tf = 0 °C - 1.12 °C
Tf ≈ -1.12 °C
Therefore, the freezing point of the solution with 0.060 mol MgSO4 in 100 g H2O is approximately -1.12 °C.