What NaCl concentration results when 204 mLof a 0.760 M NaCI solution is mixed with 417 mL of a 0.430 M NaCl solution?
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Answer:
[
N
a
C
l
]
≅
0.5
⋅
m
o
l
⋅
L
−
1
Explanation:
Solution 1
,
moles of NaCl
=
204
×
10
−
3
⋅
L
×
0.760
⋅
m
o
l
⋅
L
−
1
=
0.155
⋅
m
o
l
.
Solution 2
,
moles of NaCl
=
417
×
10
−
3
⋅
L
×
0.430
⋅
m
o
l
⋅
L
−
1
=
0.179
⋅
m
o
l
.
Since we must consider the volumes to be additive when these solutions are mixed, we have the following concentration with respect to
N
a
C
l
.
Concentration
=
(
0.155
+
0.179
)
⋅
m
o
l
(
204
+
417
)
×
10
−
3
⋅
L
≅
0.5
⋅
m
o
l
⋅
L
−
1
with respect to
N
a
C
l
.
HOPE IT HELPS YOU....
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