How many grams of potassium phosphate must be dissolved in 3.75 Liters of water to produce a soltuion that contains 8*10^23 potassium ions?
You want 8E23 K ions. That is 8E23/6.02E23 moles K ions = 1.33 moles K ions. Convert to moles K3PO4. That will be 1.33 mols K ions x (1 mol K3PO4/3 moles K ions) = 0.443 moles K3PO4.
moles K3PO4 = grams/molar mass. Solve for grams. Note: do you really care what the volume. This many K ions will this many K ions in a tub or a lake.
To determine the number of grams of potassium phosphate required, we need to use the molar mass of potassium phosphate and the Avogadro's number.
Here are the steps to calculate it:
1. Determine the molar mass of potassium phosphate (K3PO4).
- The atomic mass of potassium (K) is approximately 39.1 g/mol.
- The atomic mass of phosphorus (P) is approximately 30.97 g/mol.
- The atomic mass of oxygen (O) is approximately 16.0 g/mol.
- The molar mass of potassium phosphate (K3PO4) can be calculated as follows:
Molar mass of K3PO4 = (3 × atomic mass of K) + atomic mass of P + (4 × atomic mass of O)
2. Calculate the number of moles of potassium ions (K+):
- The given solution contains 8 × 10^23 potassium ions.
- Since each potassium phosphate molecule (K3PO4) contains three potassium ions, the number of moles of potassium ions can be calculated as:
Moles of K+ = (Number of potassium ions) / 3
3. Calculate the number of moles of potassium phosphate (K3PO4):
- Since each potassium phosphate (K3PO4) molecule contains three potassium ions, the number of moles of potassium phosphate can be calculated as:
Moles of K3PO4 = (Moles of K+) / 3
4. Calculate the mass of potassium phosphate required:
- The mass of potassium phosphate can be calculated using the following formula:
Mass = Moles of K3PO4 × Molar mass
Once you have obtained the mass of potassium phosphate, you can dissolve that amount in 3.75 Liters of water to produce the desired solution.
Please note that the exact values of atomic masses and Avogadro's number may vary slightly depending on the source used.