What is the mass of the precipitate formed when 50 mL of 16.9% solution of AgNO3 is mixed with 50 mL of 5.8% NaCl solution ?
(Ag = 107.3, N = 14, O = 16, Na = 23, Cl = 35.5)
Answers
Answer:
Well,
w/v
≡
Mass of solute
Volume of solution
, we should get over
7
⋅
g
of silver chloride.
Explanation:
And thus
mass of silver nitrate
=
16.9
%
×
50
⋅
m
L
=
8.45
⋅
g
.
And this represents a molar quantity of
8.45
⋅
g
169.87
⋅
g
⋅
m
o
l
−
1
=
0.0497
⋅
m
o
l
, with respect to
A
g
N
O
3
.
Likewise
mass of sodium chloride
=
5.8
%
×
50
⋅
m
L
=
2.90
⋅
g
.
And this represents a molar quantity of
2.90
⋅
g
58.44
⋅
g
⋅
m
o
l
−
1
=
0.0497
⋅
m
o
l
with respect to
N
a
C
l
.
Clearly the reagents are present in 1:1 molar ratio. The reaction that occurs in solution is the precipitation of a curdy white mass of
A
g
C
l
(
s
)
, i.e. and we represent this reaction by the net ionic equation.....
A
g
+
+
C
l
−
→
A
g
C
l
(
s
)
⏐
↓
Of course the complete reaction is....
A
g
N
O
3
(
a
q
)
+
N
a
C
l
(
a
q
)
→
A
g
C
l
(
s
)
⏐
↓
+
N
a
N
O
3
(
a
q
)
, i.e. sodium nitrate remains in solution and can be separated (with effort) from the precipitate.
And given the stoichiometry, we gets
0.04974
⋅
m
o
l
×
143.32
⋅
g
⋅
m
o
l
−
1
=
7.11
⋅
g
.
Of course a material such as silver halide would be very hard to isolate. Particle size is very small; it is likely to clog the filter, and filter very slowly; and moreover
A
g
C
l
is photoactive, and would decompose under light.