Question

How would you prepare 1 L of 0.04M acetic acid starting with concentrated acetic acid (density=1.05)? The manufacturer states that the concentrated acetic acid is >99.8%, so you may assume it is 100% pure.

Answer 1

M2 = 0.04 M

V2 = 1 L

density = mass / volume

since it is 100% pure for 100 mL

mass = density * volume

= 1.05 * 100

= 105g

molarity = moles / volume =(weight taken / molar mass ) / volume of solution in liter

molarity =( 105 / 60.05 ) / 0.1

molarity = 17.485 M

M1V1 = M2V2

M1 = 17.485 M

V1 = M2V2 / M1

V1 = 0.04 * 1 / 17.485

V1 = 0.00229 L

V1 = 2.29 mL

so if you take the 2.29 mL of concentrated acid ( >99.8%) and dilute it to 1000 mL ( 1 L) by adding distilled water to it you will get the 0.04 M acetic acid solution.