Solution: Here, we are going to calculate the mass of solid silver nitrate required to prepare 200.0 mL of 0.150 M silver nitrate solution. Also, we are going to calculate the volume of 3.6 M nitric acid required to prepare 100 mL of 0.50 M nitric acid solution. Step1: Molarity is defined as the number of moles of the solute(Here silver nitrate is the solute) in 1 litre of the solution. Mathematically, Number of moles of solute Molarity(M) = -------------------------------------- ------(1) Volume of solution in litres Step2: a) Given, Desired volume of the silver nitrate solution = 200 mL = 200 / 1000 L = 0.200 L Molarity of the silver nitrate solution = 0.150 M Now, using the molarity expression, we get, Number of moles of AgNO = volume o3solution in litres X Molarity AgNO3 = 0.200 X 0.150 M = 0.03 mol Now, 1 mole of AgNO = molar mass of AgNO 3 3 Therefore, 0.03 moles of AgNO = 0.03 X 3lar mass of AgNO 3 = 0.03 X 169.87 g [molar mass of AgNO = 169.87 g/mol] 3 = 5.0961...
