Answer:
Whether barium chloride solution was pure
Step-by-step explanation:
We may answer whether barium chloride was pure. The sequence of this experiment might be depicted by the following balanced chemical equations:


Having a total sample of 10.0 grams, we would firstly find the mass percentage of barium in barium chloride:
![\omega_(Ba^(2+)) = (M_(Ba))/(M_(BaCl_2)) = \frac{137.327 g/mol}{208.23 g/mol\cdot 100\% = 65.95 \%]()
This means in 10.0 g, we have a total of:
of barium cations.
The precipitate is then formed and we measure its mass. Having its mass determined, we'll firstly find the percentage of barium in barium sulfate using the same approach:

Multiplying the mass we obtained by the fraction of barium will yield mass of barium in barium sulfate. Then:
- if this number is equal to 6.595 g, we have a pure sample of barium chloride;
- if this number is lower than 6.595 g, this means we have an impure sample of barium chloride, as we were only able to precipitate a fraction of 6.595 g.