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

Mar 28, 2015

You'd need $\text{2.62 g}$ 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

$C = \frac{n}{V} \implies n = C \cdot V$

${n}_{A g N {O}_{3}} = \text{0.0260 M" * 593.4 * 10^(-3)"L" = "0.01543 moles}$

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

${\text{0.01543"cancel("moles AgNO"_3) * "169.87 g"/(cancel("1 mole AgNO"_3)) = "2.62 g AgNO}}_{3}$