Chemistry Reference
In-Depth Information
(a)
In this case, the final volume is 2.50 L, so the final concentration is
3.00
mol
1.00 L
0.625 mol
2.50 L
M
0.250 M
3.00
mol
1.00 L
(b)
In this case, the final volume is almost exactly
so the final concentration is
1.25 L
2.50 L
3.75 L,
4.00 L
3.00
mol
1.00 L
0.625 mol
3.75 L
3.00
mol
M
0.167 M
1.00 L
3.00
mol
1.00 L
Practice Problem 11.7
A nurse must prepare 4.00 L of 0.250 M saline
solution (NaCl). What volume of 6.00 M stock solution should the nurse dilute?
2.00 L
3.00
mol
1.00 L
Combining two solutions is only a little more complicated.
Figure 11.2
Combination of
Two Solutions of Equal
Concentration
EXAMPLE 11.8
Calculate the final concentration of a solution prepared by adding 2.00 L of 3.00
M sugar solution to 4.00 L of 3.00 M sugar solution.
Solution
Because the concentration of each solution is the same, the concentration of the
combined solution is also 3.00 M (Figure 11.2). (The solutions would taste as
sweet before and after mixing.)
Practice Problem 11.8
Show by calculation of the total number of
moles of sugar in the combined solution of Example 11.8 that the concentra-
tion is still 3.00 M.
EXAMPLE 11.9
Calculate the final concentration after 1.25 L of 2.25 M NaCl is added to 3.50 L
of 2.45 M NaCl and the resulting solution is diluted to 5.00 L.
Solution
The molarity of the final solution is equal to the total number of moles of solute
divided by the final volume. Solute is contained in each solution, but not in the
Liters of
solution 1
Liters of
solution 2
Mo
larit
y of
sol
utio
n 1
Mo
larit
y of
sol
utio
n 2
Moles of
solute in 1
Moles of
solute in 2
Total moles
of solute
