The idea here is that you need to figure out how many moles of magnesium chloride,
MgCl
2
, you need to have in the target solution, then use this value to determine what volume of the stock solution would contain this many moles.
As you know, molarity is defined as the number of moles of solute, which in your case is magnesium chloride, divided by liters of solution.
c
=
n
V
So, how many moles of magnesium chloride must be present in the target solution?
c
=
n
V
⇒
n
=
c
⋅
V
n
=
0.158 M
⋅
250.0
⋅
10
−
3
L
=
0.0395 moles MgCl
2
Now determine what volume of the target solution would contain this many moles of magnesium chloride
c
=
n
V
⇒
V
=
n
c
V
=
0.0395
moles
3.15
moles
L
=
0.01254 L
Rounded to three sig figs and expressed in mililiters, the volume will be
V
=
12.5 mL
So, to prepare your target solution, use a
12.5-mL
sample of the stock solution and add enough water to make the volume of the total solution equal to
250.0 mL
.
This is equivalent to diluting the
12.5-mL
sample of the stock solution by a dilution factor of
20
.
Free energy is the answer i hope this helped
Answer:
1. 1.25 mol ants x 6.02*10^23 ants/1 mol ants = 7.53*10^23 ants
2. 4.92*10^26 pencils x 1 mol pencils/6.02*10^23 pencils = 817 mol pencils
3. 0.26 mol molecules x 6.02*10^23 molecules/1 mol molecules = 1.6*10^23 molecules
4. 3.46*10^19 molecules x 1 mol molecules/6.02*10^23 molecules = 5.75*10^-5 mol molecules
5. 5.3*10^20 atoms x 1 mol atoms/6.02*10^23 atoms = 8.8 mol atoms
6. 0.11 mol atoms x 6.02*10^23 atoms/1 mol atoms = 6.6*10^22 atoms
I would suggest looking into "dimensional analysis" for help with this type of material. Dimensional analysis will stick with you all throughout chemistry, so picking it up will be extremely beneficial.
