Question

Compute the number of ions present in 5.85g of sodium chlorude

Answer 1

Answer:

5.85 g ofNaCl = 1 X 5.85/58.5 = 0.1 mole. Each molecule of NaCI contains one Na ion and one Cl ion = 2 ions. Total moles of ions in 0.1 mole of NaCI = 2 x 0.1 = 0.2 moles.

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Answer 2

1 mole of NaCl = 58.5 g

x moles of NaCl = 5.85 g

By cross-multiplication , we get

5.85/58.5 = 0.1 moles

1 NaCl contains 1 Na+ ion and 1 Cl- ion, a total of 2 ions.

Thus 1 mole of NaCl contains 2*(Avogadro Number) of ions

=> 0.1 moles contains 0.2*(Avogadro Number) ions = 1.2 x 10^23 ions