What is the freezing pint of 0.0091 m aqueous sodium phosphate, Na3PO4? Assume that sodium phosphate is a strong electrolyte and it functions ideally in solution. (express your answer in degrees Celsius)
I do believe that
Van Hoff = 4
Kf(water) = 1.86
Molality = .0091
4*1.86*0.0091=delta T=.067704
0.00-.067704=-0.0677 degrees C is the answer on Millers quiz
To find the freezing point of a solution, we need to use the formula for freezing point depression:
ΔT = Kf * m * i
Where:
ΔT = change in freezing point
Kf = cryoscopic constant (specific to the solvent)
m = molality of the solute
i = van't Hoff factor (number of particles the solute dissociates into when it dissolves)
For water (the solvent), the cryoscopic constant (Kf) is approximately 1.86 °C/m.
In this case, we have an aqueous solution of sodium phosphate, Na3PO4, which is a strong electrolyte. When it dissociates in water, it forms four ions: 3 Na+ ions and 1 PO4^-3 ion. So the van't Hoff factor (i) for sodium phosphate is 4.
The concentration of the solution can be expressed in molality (moles of solute per kilogram of solvent). To calculate molality, we need to know the molar mass of sodium phosphate.
The molar mass of Na3PO4 is:
(3 * molar mass of Na) + molar mass of P + (4 * molar mass of O)
= (3 * 22.99 g/mol) + 30.97 g/mol + (4 * 16.00 g/mol)
= 163.94 g/mol
Now, we can calculate the molality (m):
molality (m) = moles of solute / mass of solvent (in kg)
To find the moles of sodium phosphate:
moles of Na3PO4 = concentration (in mol/L) * volume of solution (in L)
Given that the concentration is 0.0091 m and assuming a volume of 1 L for simplicity, we can calculate the moles of sodium phosphate as:
moles of Na3PO4 = 0.0091 mol/L * 1 L = 0.0091 mol
Next, we need to determine the mass of the solvent (water) in the solution. Assuming a density of 1.00 g/mL for water, the mass of 1 L of water is 1000 g (or 1 kg).
Now we can calculate the molality (m):
molality (m) = 0.0091 mol / 1 kg = 0.0091 mol/kg
Finally, we can use the formula for freezing point depression to find the change in freezing point (ΔT) and then subtract it from the normal freezing point of water (0 °C) to find the freezing point of the solution:
ΔT = Kf * m * i
ΔT = 1.86 °C/m * 0.0091 mol/kg * 4
ΔT = 0.067 °C
Freezing point of the solution = Freezing point of water - ΔT
Freezing point of the solution = 0 °C - 0.067 °C = -0.067 °C
Therefore, the freezing point of the 0.0091 m aqueous sodium phosphate solution is approximately -0.067 °C.