1. Find the molar mass of AgNO3:
* Ag (Silver): 107.87 g/mol
* N (Nitrogen): 14.01 g/mol
* O (Oxygen): 16.00 g/mol (x 3 = 48.00 g/mol)
* Total molar mass of AgNO3 = 107.87 + 14.01 + 48.00 = 169.88 g/mol
2. Convert the mass of AgNO3 to moles:
* Moles of AgNO3 = (42.5 g) / (169.88 g/mol) = 0.250 mol
3. Convert the volume of the solution to liters:
* 100 mL = 0.100 L
4. Calculate the molarity (M):
* Molarity (M) = (Moles of solute) / (Liters of solution)
* M = (0.250 mol) / (0.100 L) = 2.50 M
Therefore, the molarity of the silver nitrate solution is 2.50 M.
