What is the molarity of a salt solution that is made from 31.0 grams of Ca3(P04)2 placed in a volumetric flask and filled to the 2 liter line with distilled water?

I don't know if this is a REAL problem or not? If it is just a made up problem by someone who has slipped up, the answer (not real) is

mols = grams/molar mass, then
M = mols/2L = ?

However, if you recognize that Ca3(PO4)2 is an insoluble salt in H2O, then you get a REAL answer this way.
........Ca3(PO4)2 ==> 3Ca^2+ + 2PO4^3-
I........solid.........0........0
C........solid.........3x.......2x
E........solid.........3x.......2x

Ksp = (Ca^2+)^3(PO4^3-)^2
Ksp = 2.07E-33
2.07E-33 = (3x)^3(2x)^2
2.07E-33 = 27x^3*4x^2 = 108x^5
Solve for x.
x is the REAL molarity of Ca3(PO4)2 in water solution.

To determine the molarity of the salt solution, we need to follow a few steps:

1. Calculate the molar mass of Ca3(PO4)2 (Calcium phosphate).
The molar mass of Ca is 40.08 g/mol, P is 30.97 g/mol, and O is 16.00 g/mol.

Molar mass of Ca3(PO4)2 = (3 * 40.08) + (2 * 30.97) + (8 * 16.00) g/mol
= 120.24 + 61.94 + 128.00 g/mol
= 310.18 g/mol

2. Convert the given mass of Ca3(PO4)2 to moles.
Moles = Mass / Molar mass

Moles of Ca3(PO4)2 = 31.0 g / 310.18 g/mol
= 0.09992 mol

3. Calculate the molarity (M) of the solution.
Molarity (M) = Moles of solute / Volume of solution (in liters)

Since the solution is made in a 2 liter volumetric flask, the volume of solution is 2.0 L.

Molarity (M) = 0.09992 mol / 2.0 L
= 0.04996 M

Therefore, the molarity of the salt solution is 0.04996 M.

15.5 M