How to solve molality problem?

Question

In a lab additional questions it asks,
Describe how to prepare 250.0 ml of 0.1000 M AgNO3 from pure solid silver nitrate.

I don't understand how to solve it. Someone please explain.

Answer 1

0.1 M AgNO3 means 0.1 moles AgNO3 in 1 L solution. You only need 250 mL (0.25 L).

Molarity is # moles solute in 1 L solution. You need 0.1 M AgNO3

(0.1 mol AgNO3 / 1000 mL ) = (? mol / 250 mL)
This gives you the number of moles needed. Then multiply by molecular mass of AgNO3 to get grams needed to add to 250 mL water.