# How many grams of silver nitrate are needed to prepare 250 mL of standard 0.100 M silver nitrate solution?

Jun 7, 2017

${\text{4.2 g AgNO}}_{3}$

#### Explanation:

For starters, you know that a solution's molarity tells you the number of moles of solute present for every $\text{1 L}$ of solution.

In your case, you have

"molarity" = color(blue)("0.100 mol")color(white)(.)color(darkorange)("L"^(-1))

This means that for every

$\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\text{1 L") = color(darkorange)(10^3color(white)(.)"mL}}$

your solution will contain $\textcolor{b l u e}{\text{0.100 moles}}$ of silver nitrate, ${\text{AgNO}}_{3}$. the solute.

Now, notice that your solution has a volume of

"250 mL" = (color(darkorange)(10^3color(white)(.)"mL"))/color(red)(4)

This means that in order to have a molarity of ${\text{0.100 mol L}}^{- 1}$, the solution must contain

color(blue)("0.100 moles AgNO"_3)/color(red)(4) = "0.025 moles AgNO"_3

Finally, to convert the number of moles to grams, use the molar mass of silver nitrate

$0.025 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles AgNO"_3))) * "169.87 g"/(1color(red)(cancel(color(black)("mole AgNO"_3)))) = color(darkgreen)(ul(color(black)("4.2 g}}}}$

The answer is rounded to two sig figs, the number of sig figs you have for the volume of the solution.