What is the molarity of a solution prepared by diluting 100g of NaH2PO4 to 1L?
100 g/L = ??mols/L.
mols = g/molar mas.
To determine the molarity of a solution, we need to know the number of moles of solute and the volume of the solution.
First, let's find the number of moles of NaH2PO4 in 100g.
The molar mass of NaH2PO4 can be calculated by adding up the atomic masses of its constituent elements:
Na: 22.99 g/mol
H: 1.01 g/mol (there are 2 hydrogen atoms in NaH2PO4)
P: 30.97 g/mol
O: 16.00 g/mol (there are 4 oxygen atoms in NaH2PO4)
Molar mass of NaH2PO4 = (22.99 g/mol) + (2 * 1.01 g/mol) + 30.97 g/mol + (4 * 16.00 g/mol)
= 119.98 g/mol
Next, we need to convert the mass of NaH2PO4 to moles using the equation:
moles = mass / molar mass
moles = 100g / 119.98 g/mol
moles ≈ 0.833 mol
Now, to find the molarity of the solution, we need to know the volume. In this case, the volume is given as 1L (or 1000 mL).
Finally, we can calculate the molarity using the equation:
Molarity (M) = moles of solute / volume of solution (in L)
Molarity = 0.833 mol / 1 L
Molarity = 0.833 M
Therefore, the molarity of the solution prepared by diluting 100g of NaH2PO4 to 1L is approximately 0.833 M.