the solubility of Ag3PO4 (in water) is 6.7 x 10^-3 g/L. what is the molar solubility>

moles = grams/molar mass

To determine the molar solubility of Ag3PO4, we need to convert the given solubility from grams per liter (g/L) to moles per liter (mol/L).

First, let's calculate the molar mass of Ag3PO4. The atomic mass of silver (Ag) is 107.87 g/mol, and the atomic mass of phosphorus (P) is 30.97 g/mol. Oxygen (O) has an atomic mass of 16.00 g/mol.

Molar mass of Ag3PO4:
(3 × atomic mass of Ag) + atomic mass of P + (4 × atomic mass of O)
= (3 × 107.87 g/mol) + 30.97 g/mol + (4 × 16.00 g/mol)
≈ 418.57 g/mol

Now, we will convert the solubility from grams per liter (g/L) to moles per liter (mol/L).

Given solubility: 6.7 × 10^(-3) g/L

First, convert grams to moles by using the molar mass of Ag3PO4:
6.7 × 10^(-3) g × (1 mol / 418.57 g) = 1.601 × 10^(-5) mol

Next, divide the number of moles by the volume in liters:
1.601 × 10^(-5) mol / 1 L ≈ 1.601 × 10^(-5) mol/L

Therefore, the molar solubility of Ag3PO4 is approximately 1.601 × 10^(-5) mol/L.