M=mol/L
use stoichiometry
0.160M=0.160mol/L
find mol
0.160mol/L*0.240L=0.0384 mol
multiply by molar mass of silver nitrate
0.0384mol*169.87g/mol=6.523g AgNO3
How many grams of solid silver nitrate would you need to prepare 240.0mL of a 0.160M AgNO3 solution?
1 answer