An aqueous solutions is 18.7% by mass Na3PO4. What is the boiling point of this solution? The Kb for water is .512 C/m. Consider if the solute is an electrolyte or a nonelectrolyte.
18.7% means 18.7 g/100 g SOLUTION. Converted to 1000g solution that is 187g/1000 g solution. How much of that is H2O? 1000-187 = about 813g.
mols Na3PO4 = 187/molar mass = about 1.1 mols. So m = mols/kg soln or
m = 1.1/0.8 = about 1.4 or so. You need to go through these calculations and do them more accurately.
Then delta T = i*Kb*m
i = 4
Kb you have
m = your value for m
To find the new boiling point added delta T to 100 C.
To find the boiling point of the solution, we need to consider the colligative properties of the solute. Colligative properties depend on the number of solute particles rather than the nature of the solute itself.
Since Na3PO4 is ionic and dissociates into ions when dissolved in water, it is considered an electrolyte. When an electrolyte dissolves in water, it forms multiple particles in solution, increasing the effect on the boiling point.
To calculate the boiling point elevation, we can use the formula:
ΔTb = Kb * m * i
Where:
ΔTb is the boiling point elevation,
Kb is the molal boiling point elevation constant for water (0.512 °C/m),
m is the molality of the solution (moles of solute/kg of solvent),
i is the van't Hoff factor, which represents the number of particles into which each formula unit or molecule of solute dissociates.
In the case of Na3PO4, it dissociates into four ions:
Na3PO4 → 3Na+ + PO4^3-
So, the van't Hoff factor (i) for Na3PO4 is 4.
To find the molality (m) of the solution, we need to determine the amount of Na3PO4 in moles and the mass of water.
Let's assume we have 100g of the solution.
So, the mass of Na3PO4 = 18.7% of 100g = 18.7g
To convert grams to moles, we need to know the molar mass of Na3PO4:
Na (sodium) = 22.99 g/mol
P (phosphorus) = 30.97 g/mol
O (oxygen) = 16.00 g/mol
Molar mass of Na3PO4 = (3 * 22.99) + 30.97 + (4 * 16.00) = 163.94 g/mol
Now, we can calculate the moles of Na3PO4:
moles of Na3PO4 = mass / molar mass = 18.7g / 163.94 g/mol
To find the molality (m), we need to determine the mass of water.
Mass of water = total mass of solution - mass of Na3PO4 = 100g - 18.7g
Now, we can calculate the molality:
m = moles of Na3PO4 / mass of water
Finally, we can substitute the values into the boiling point elevation formula to find ΔTb, the boiling point elevation. Adding ΔTb to the normal boiling point of water (100°C) will give us the boiling point of the solution.
I hope this clarifies the process! Let me know if you need any further assistance.