Question

you are ask to prepare 125.0mL of 0.0321M AgNO3. How many grams would you need of a sample known to be 99.81% AgNO3 by mass?

Answer 1

You're given a concentration and volume of solution to make. That info gets you moles since:\[Conc = (\frac{mol}{Liter})\]then \[mol = Conc * Liters\]
Now you have moles of solute that must be added to enough water to make the given volume of solution. Use the molar mass of the solute:\[(mol AgNO{_3})*(\frac{169.8g AgNO{_3}}{1mol AgNO{_3}})\] should get you 0.681g IF the sample were 100% pure. Since it's close, but not quite pure, you will need:\[0.681g* (\frac{100\%}{99.81\%})\] or 0.683g.