Here's what I got.
Explanation:
Interestingly enough, I'm not getting
0.0341% w/v
either. Here's why.
Start by calculating the percent composition of chlorine,
Cl
, in calcium chloride, This will help you calculate the mass of chloride anions,
Cl
−
, present in your sample.
To do that, use the molar mass of calcium chloride, the molar mass of elemental chlorine, and the fact that
1
mole of calcium chloride contains
2
moles of chlorine atoms.
2
×
35.453
g mol
−
1
110.98
g mol
−
1
⋅
100
%
=
63.89% Cl
This means that for every
100 g
of calcium chloride, you get
63.89 g
of chlorine.
As you know, the mass of an ion is approximately equal to the mass of the neutral atom, so you can say that for every
100 g
of calcium chloride, you get
63.89 g
of chloride anions,
Cl
−
.
This implies that your sample contains
0.543
g CaCl
2
⋅
63.89 g Cl
−
100
g CaCl
2
=
0.3469 g Cl
−
Now, in order to find the mass by volume percent concentration of chloride anions in the resulting solution, you must determine the mass of chloride anions present in
100 mL
of this solution.
Since you know that
500 mL
of solution contain
0.3469 g
of chloride anions, you can say that
100 mL
of solution will contain
100
mL solution
⋅
0.3469 g Cl
−
500
mL solution
=
0.06938 g Cl
−
Therefore, you can say that the mass by volume percent concentration of chloride anions will be
% m/v = 0.069% Cl
−
−−−−−−−−−−−−−−−−−−−
I'll leave the answer rounded to two sig figs, but keep in mind that you have one significant figure for the volume of the solution.
.
ALTERNATIVE APPROACH
Alternatively, you can start by calculating the number of moles of calcium chloride present in your sample
0.543
g
⋅
1 mole CaCl
2
110.98
g
=
0.004893 moles CaCl
2
To find the molarity of this solution, calculate the number of moles of calcium chloride present in
1 L
=
10
3
mL
of solution by using the fact that you have
0.004893
moles present in
500 mL
of solution.
10
3
mL solution
⋅
0.004893 moles CaCl
2
500
mL solution
=
0.009786 moles CaCl
2
You can thus say your solution has
[
CaCl
2
]
=
0.009786 mol L
−
1
Since every mole of calcium chloride delivers
2
moles of chloride anions to the solution, you can say that you have
[
Cl
−
]
=
2
⋅
0.009786 mol L
−
1
[
Cl
−
]
=
0.01957 mol L
−
This implies that
100 mL
of this solution will contain
100
mL solution
⋅
0.01957 moles Cl
−
10
3
mL solution
=
0.001957 moles Cl
−
Finally, to convert this to grams, use the molar mass of elemental chlorine
0.001957
moles Cl
−
⋅
35.453 g
1
mole Cl
−
=
0.06938 g Cl
−
Once again, you have
% m/v = 0.069% Cl
−
−−−−−−−−−−−−−−−−−−−
In reference to the explanation you provided, you have
0.341 g L
−
1
=
0.0341 g/100 mL
=
0.0341% m/v
because you have
1 L
=
10
3
mL
.
However, this solution does not contain
0.341 g
of chloride anions in
1 L
. Using
[
Cl
−
]
=
0.01957 mol L
−
1
you have
n
=
c
⋅
V
so
n
=
0.01957 mol
⋅
10
−
3
mL
−
1
⋅
500
mL
n
=
0.009785 moles
This is how many moles of chloride anions you have in
500 mL
of solution. Consequently,
100 mL
of solution will contain
100
mL solution
⋅
0.009785 moles Cl
−
500
mL solution
=
0.001957 moles Cl
−
So once again, you have
0.06938 g
of chloride anions in
100 mL
of solution, the equivalent of
0.069% m/v
.
Explanation:
i think this is it
