How many grams of AgNO3 are needed to prepare 593.4 mL of 0.0260 M AgNO3 solution?

1 Answer
Stefan V.
Mar 28, 2015

You'd need #"2.62 g"# of silver nitrate to prepare that particular solution.

Again, molarity is defined as moles of solute per liter of solution. You know the solution's molarity and its volume, which means you can solve for the number of moles of silver nitrate

#C = n/V => n = C * V#

#n_(AgNO_3) = "0.0260 M" * 593.4 * 10^(-3)"L" = "0.01543 moles"#

Use silver nitrate's molar mass to solve for the mass in grams

#"0.01543"cancel("moles AgNO"_3) * "169.87 g"/(cancel("1 mole AgNO"_3)) = "2.62 g AgNO"_3#

Related questions
See all questions in Molarity
Impact of this question
23738 views around the world
You can reuse this answer
Creative Commons License