Knowing that the density of a 17.4M CH3COOH
solution = 1.05148 g/mL
a) How many mL of a 17.4M a acetic acid stock solution
is needed for making 3500 mL of a 2M acetic acid solution.
b) How many grams of a 17.4M a acetic acid stock
solution is needed for making 3500 mL of a 2M acetic acid
solution.
We know that:
M = n/Vsolution
C1 V1
= C2 V2
(dilution formula)
Solution a)
17.4M * C2 = 3.5L
* 2M
V2 = (3.5L * 2M)/17.4M =
0.4023 L = 402.3mL
Solution b)
Since we already know (from solution a) )
the volume of a 17.4M solution needed and the density
we can calculate the mass :
We need to measure out 402.3mL * 1.05148 g/mL=
423.01g of 17.4M acetic acid