how many moles of sodium ions are present in 3.50 L of 0.400 M Na3PO4?
How many moles Na3PO4 are present? That is moles = M x L.
Then moles Na must be just 3 times that since there re 3 Na ions/ molecule of Na3PO4.
To determine the number of moles of sodium ions in 3.50 L of a 0.400 M Na3PO4 solution, we need to use the relationship between moles, concentration, and volume.
The Na3PO4 compound contains 3 sodium ions (Na+) per molecule. This means that for every mole of Na3PO4, there are 3 moles of sodium ions.
First, let's calculate the number of moles of Na3PO4 in the given solution:
Molarity (M) = moles of solute / liters of solution
0.400 M = moles of Na3PO4 / 3.50 L
Rearranging the equation, we get:
moles of Na3PO4 = 0.400 M * 3.50 L
moles of Na3PO4 = 1.40 moles
Since there are 3 moles of sodium ions (Na+) per mole of Na3PO4, we can calculate the number of moles of sodium ions:
moles of Na+ = 3 moles of Na3PO4 * (1 mole of Na+ / 1 mole of Na3PO4)
moles of Na+ = 3 * 1.40 moles
moles of Na+ = 4.20 moles
Therefore, there are 4.20 moles of sodium ions present in 3.50 L of the 0.400 M Na3PO4 solution.