Answer:
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.
.
Explanation:
