Task:
How many Mg ions are present in 3.00 moles of MgCl2?
Solution:
In water MgCl2 dissociates to ions: a cation of magnesium (Mg2+) and two anions of cloride (Cl-).
MgCl2 = Mg2+ + 2Cl-
Hence, n(MgCl2) = n(Mg2+) = 3.00 moles (according to the reaction equation).
n = N / Na, where Na - Avogadro's number = 6.022*1023 mol-1.
Then,
N(Mg2+) = n(Mg2+) * Na = 3.00 * 6.022*1023 = 18.066*1023 = 1.81*1024
Answer: 1.81*1024 Mg2+ ions are present in 3.00 moles of MgCl2.
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