how many grams of solid silver nitrate would you need to prepare 200.0 mL of a 0.150 M AgNO3 solution
1 answer:
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Explanation:
The given data is as follows.
= 225.5 mL,
= 45.00 mL
= 0.09988 M,
= ?
Therefore, formula to calculate final concentration will be as follows.
=
Putting the given values into the above formula as follows.
=
=
= 0.5 M
Thus, we can conclude that the final concentration of the given solution is 0.5 M.