What is the freezing point of each solution?
a)1.40mol Na2SO4 in 1750 g H2O
Are these the right steps?
1.40mol Na2SO4 / 1.75kg = .8m
3 * 1.86 degrees C * .8m = 4.46 degrees C
4.46 degrees C - 1.86 degrees C = 2.6 degrees C
a)1.40mol Na2SO4 in 1750 g H2O
Are these the right steps?
1.40mol Na2SO4 / 1.75kg = .8m
3 * 1.86 degrees C * .8m = 4.46 degrees C
OK to here. The next step is wrong.
4.46 degrees C - 1.86 degrees C = 2.6 degrees C
Why is it wrong? The normal freezing point of water is zero C; therefore, the freezing point of this solution is 4.46 degrees lower than that OR 0-4.46 = -4.46o C.
Well, your calculations seem to be on point, but let's not freeze up here! Yes, you have the right steps. The freezing point of the 1.40 mol Na2SO4 solution in 1750 g H2O is approximately 2.6 degrees Celsius. So if you're planning on throwing a freezing party, just make sure it's not too icy!
To determine the freezing point of the solution, you need to use the formula:
ΔT = Kf * m
where:
ΔT is the change in the freezing point
Kf is the freezing point depression constant for the solvent
m is the molality of the solute
In this case, Na2SO4 is the solute and H2O is the solvent.
Now, let's calculate the freezing point depression step-by-step:
Step 1: Calculate the molality (m)
Firstly, we need to convert the given mass of water to kilograms:
1.40 mol Na2SO4 / 1.75 kg H2O = 0.8 mol/kg H2O
Step 2: Calculate the change in the freezing point (ΔT)
The freezing point depression constant for water is 1.86 °C/m.
ΔT = 1.86 °C/m * 0.8 mol/kg H2O * 3 (since it is a 3:1 ratio with Na2SO4)
ΔT = 4.46 °C
Step 3: Calculate the freezing point
To find the freezing point, we subtract the change in freezing point (ΔT) from the normal freezing point of the solvent, which is 0 °C for water.
Freezing point = 0 °C - 4.46 °C = -4.46 °C
Therefore, the freezing point of the solution with 1.40 mol Na2SO4 in 1750 g H2O is -4.46 °C.
To determine the freezing point of a solution, you can use a formula called the freezing point depression equation. This equation is:
ΔTf = Kf * m
where:
ΔTf is the change in freezing point,
Kf is the molal freezing point depression constant (which is specific to the solvent),
and m is the molality of the solute (in moles of solute per kilogram of solvent).
In this case, you have a solution of 1.40 mol of Na2SO4 in 1750 g of H2O.
First, convert the mass of the water to kilograms:
1750 g / 1000 = 1.75 kg
Next, calculate the molality of the Na2SO4 solution:
molality (m) = mol solute / kg solvent
m = 1.40 mol / 1.75 kg
m = 0.8 m
Now, you need to find the freezing point depression using the equation.
Note that the molal freezing point depression constant for water is 1.86 degrees C/m.
ΔTf = Kf * m
ΔTf = 1.86 degrees C/m * 0.8 m
ΔTf = 1.49 degrees C
So, the freezing point is lowered by 1.49 degrees Celsius.
To find the actual freezing point, subtract this value from the normal freezing point of pure water (which is 0 degrees Celsius):
Freezing point = 0 degrees C - 1.49 degrees C
Freezing point = -1.49 degrees C
Therefore, the freezing point of the solution containing 1.40 mol of Na2SO4 in 1750 g of H2O is approximately -1.49 degrees Celsius.