Calculate the mass of sodium phosphate required to prepare 1.75 L of solution in which the sodium ion concentration is 0.25 mol/L
I believe the answer provided by this solution is three times too large. The problem asks for grams Na3PO4 to provide 0.25 M SODIUM ION. Since there are 3 Na ions per Na3PO4 molecule you need to divide 0.25/3 to begin with, then proceed s in the above.
The units given are in mol/L or M
Solve for moles:
0.25mol/L= x moles/1.75L
Solve for x moles:
x moles=1.75L*(0.25mol/L)
x moles * molecular weight of sodium phosphate (163.94 g/mol)= mass or Sodium Phosphate
No more than two significant figures in answer.
Well, let's calculate it in a *sodium*arily ridiculous way! First, we need to determine the number of moles of sodium ions present in the solution.
The sodium ion concentration is given as 0.25 mol/L. To find out how many moles are present in 1.75 L, we can multiply the concentration by the volume:
0.25 mol/L x 1.75 L = 0.4375 mol
Now, we need to find the molar mass of sodium phosphate (Na3PO4) to calculate the mass. The molar mass of sodium is approximately 22.99 g/mol, and the molar mass of phosphorous is around 30.97 g/mol. Oxygen has a molar mass of approximately 16.00 g/mol.
Adding them all up, we get:
(22.99 g/mol x 3) + (30.97 g/mol) + (16.00 g/mol x 4) = 163.97 g/mol
To find the mass of sodium phosphate required, we can multiply the number of moles by the molar mass:
0.4375 mol x 163.97 g/mol ≈ 71.62 g
So, approximately 71.62 grams of sodium phosphate would be needed to prepare 1.75 L of the solution with a sodium ion concentration of 0.25 mol/L.
To calculate the mass of sodium phosphate required to prepare a solution, you'll need to know the molar mass of sodium phosphate and the desired concentration of sodium ions.
The chemical formula for sodium phosphate is Na3PO4.
Step 1: Calculate the molar mass of sodium phosphate.
The molar mass of sodium (Na) is approximately 23 g/mol.
The molar mass of phosphorus (P) is approximately 31 g/mol.
The molar mass of oxygen (O) is approximately 16 g/mol.
So, the molar mass of sodium phosphate (Na3PO4) can be calculated as:
(3 x 23 g/mol) + (1 x 31 g/mol) + (4 x 16 g/mol) = 163 g/mol.
Step 2: Calculate the number of moles of sodium ions needed for the desired concentration.
The given concentration is 0.25 mol/L. Since sodium phosphate contains three sodium ions (Na+), we need to multiply the concentration by 3 to get the sodium ion concentration.
0.25 mol/L x 3 = 0.75 mol/L.
Step 3: Calculate the mass of sodium phosphate.
The mass can be calculated using the formula:
Mass = moles x molar mass.
Mass = 0.75 mol/L x 1.75 L x 163 g/mol = 213.75 grams.
Therefore, the mass of sodium phosphate required to prepare a 1.75 L solution with a sodium ion concentration of 0.25 mol/L is approximately 213.75 grams.
To calculate the mass of sodium phosphate required to prepare a solution, you will need to know the molar mass of sodium phosphate.
The formula for sodium phosphate is Na3PO4. To calculate its molar mass, you can look up the atomic masses of sodium (Na), phosphorus (P), and oxygen (O) in the periodic table.
Na: 22.99 g/mol
P: 30.97 g/mol
O: 16.00 g/mol
Next, you need to add up the molar mass of each atom in the compound:
(3 x Na) + P + (4 x O) = (3 x 22.99) + 30.97 + (4 x 16.00) = 163.94 g/mol
Now, you have the molar mass of sodium phosphate, which is 163.94 g/mol.
The concentration of sodium ions in the solution is given as 0.25 mol/L. This means that for every liter of solution, there are 0.25 moles of sodium ions.
Since you want to prepare 1.75 L of solution, you can multiply the concentration by the volume:
0.25 mol/L x 1.75 L = 0.4375 moles
To find the mass of sodium phosphate required, you can use the equation:
mass = moles x molar mass
mass = 0.4375 moles x 163.94 g/mol = 71.61 g
So, to prepare 1.75 L of solution with a sodium ion concentration of 0.25 mol/L, you would need 71.61 grams of sodium phosphate.