I want to just know how to set this problem up and why i am having alot of trouble with it?

Question

A student found that there are 113 drops of glacial acetic acid (100% pure acetic acid) in 2.00 mL. Calculate the number of drops of glacial acetic acid, HC2H3O2 required to make 285 mL of 0.05 M acetic acid solution

Answer 1

First of all calculate how much glacial acetic you need to make the 285 ml of the 0.05M soln

0.05 moles/liter = x moles/.285liter

xmoles = 0.01425moles

CH3COOH MW = 60 g/mole , so you need

0.01425 moles * (60g/mole) = 0.85 g

The density of acetic acid is 1.049 g/ml

so you need 0.85 g * ml/1.049g = 0.81 ml

You know 113 drops = 2 ml or 113drops/2ml

so Finally

you get to it:

0.81ml *113 drops/2ml ≈ 46 drops

Answer 2

You're having a problem because you need to know the mass of pure acid, so you need to know the density of the stuff. Then you can work out what mass is is each drop, and how many moles (using the Mr as well) you will need to make the solution specified.