What is the molality of ammonium ions in a 1.75 m solution of (NH4)3PO4?
To find the molality of ammonium ions in a solution, we'll need to know the moles of ammonium ions and the mass of the solvent.
1. Calculate the moles of ammonium ions:
- The (NH4)3PO4 compound dissociates into 3 ammonium ions, (NH4)+, for every formula unit in the solution.
- So, the moles of ammonium ions can be calculated by multiplying the molarity of the solution by the number of ammonium ions produced by each mole of (NH4)3PO4.
Moles of ammonium ions = (molarity) × (number of ammonium ions produced by each mole of (NH4)3PO4)
= 1.75 mol/L × 3 mol/(1 mol)
2. Determine the mass of the solvent:
- The molality of a solution is defined as the moles of solute per kilogram of solvent.
- Since we are given the mass of the solution, we need to convert it to the mass of the solvent by subtracting the mass of the solute (ammonium ions).
Mass of solvent = Mass of solution - Mass of solute
Note: In this case, the solvent is water (H2O).
3. Finally, calculate the molality of ammonium ions:
Molality = Moles of ammonium ions / Mass of solvent
By following these steps, you will be able to calculate the molality of ammonium ions in the given solution.
To determine the molality of ammonium ions in a solution of (NH4)3PO4, we need to first understand what molality is.
Molality is a measurement of the concentration of a solute in a solution. It is defined as the number of moles of solute per kilogram of solvent. It is denoted by the symbol 'm'.
In this case, the solute is the ammonium ions (NH4+) in the compound (NH4)3PO4, and the solvent is the solution itself.
To find the molality, we need to calculate the number of moles of the ammonium ions and the mass of the solvent.
The molar mass of (NH4)3PO4 can be calculated as follows:
1 atom of N (14.01 g/mol) + 12 atoms of H (1.01 g/mol) + 3 atoms of P (30.97 g/mol) + 4 atoms of O (16.00 g/mol) = 149.09 g/mol
Since we have a 1.75 m solution, the given molality is 1.75 mol/kg.
Now, we need to calculate the amount of ammonium ions present in the solution.
The (NH4)3PO4 salt dissociates into 3 NH4+ ions and 1 PO4-3 ion.
Hence, the number of moles of ammonium ions is:
1.75 mol/kg x 3 (moles of NH4+ in (NH4)3PO4) = 5.25 mol/kg
Therefore, the molality of ammonium ions in a 1.75 m solution of (NH4)3PO4 is 5.25 mol/kg.
1 (NH4)3PO4 = 3 NH4 + 1 PO4
You need to strat from the balanced equation for the solution of (NH4)3PO4 in water to give the ions.
(NH4)3PO4 -->