Answer:
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to calculate the volume of the new solution by using the general formula of dilution:
In such a way, we solve for the final volume, V2, to obtain:
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The given question is incomplete. The complete question is as follows.
A solution contains an unknown mass of dissolved barium ions. When sodium sulfate is added to the solution, a white precipitate forms. The precipitate is filtered and dried and then found to have a mass of 212 mg. What mass of barium was in the original solution? (Assume that all of the barium was precipitated out of solution by the reaction.)
Explanation:
When and
are added then white precipitate forms. And, reaction equation for this is as follows.
It is given that mass (m) is 212 mg or 0.212 g (as 1 g = 1000 mg). Molecular weight of is 233.43.
Now, we will calculate the number of moles as follows.
No. of moles = mass × M.W
=
= 0.00091 mol of
Hence, it means that 0.00091 mol of . Now, we will calculate the mass as follows.
Mass = moles × MW
=
= 0.124 grams or 124 mg of barium
Thus, we can conclude that mass of barium into the original solution is 124 mg.