Answer:
0.6941 mg
Explanation:
First we calculate how many LiNO₃ moles there are, using the given concentration and volume:
- 1.0 mL * 0.10 M = 0.10 mmol LiNO₃
As 1 mol of LiNO₃ contains 1 mol of Li, in the problem solution there are 0.10 mmol of Li (the only metallic ion present).
Now we convert Li milimoles into miligrams, using its atomic mass:
- 0.10 mmol Li * 6.941 mg/mmol = 0.6941 mg
