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Explanation:
You can solve this problem by using the dilution factor,
DF
, which essentially tells you the factor by which the concentration of an initial solution decreased upon dilution.
The dilution factor is calculated by using either volumes or molarities
So, the dilution factor for the first dilution
58.0 mL
→
248 mL
will be
DF
1
=
248
mL
58.0
mL
=
4.276
The first solution was diluted by a factor of
4.276
, which means that its concentration decreased by a factor of
4.276
. Therefore, the concentration of the diluted solution will be
c
diluted
=
c
concentrated
DF
1
c
diluted
=
1.30 M
4.276
=
0.304 M
Now, you take a sample of
124 mL
of this diluted solution and add another
165 mL
of water. The final volume of the solution will be
V
final
=
124 mL
+
165 mL
=
289 mL
The concentration of the
124 mL
sample is equal to the concentration of the first diluted solution, i.e.
0.304 M
.
The dilution factor for the second dilution will be
DF
2
=
V
final
V
sample
DF
2
=
289
mL
124
mL
=
2.331
Therefore, the concentration of the final solution will be
c
final
=
c
sample
DF
2
c
final
=
0.304 M
2.331
=
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
0.130 M
a
a
∣
∣
−−−−−−−−−−−
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