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

For starters, you know that a solution's molarity tells you the number of moles of solute present for every #"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

#color(darkorange)("1 L") = color(darkorange)(10^3color(white)(.)"mL")#

your solution will contain #color(blue)("0.100 moles")# of silver nitrate, #"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 #"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 color(red)(cancel(color(black)("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.