Q38Long answer
Compute the number of ions present in 5.85 g of sodium chloride.
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5.85 g of NaCl = = 0.1moles SVLS or 0.1 moles of NaCl particle Each NaCl particle is equivalent to one Na+ one Cl– ⇒ 2 ions Total moles of ions = 0.1 × 2 0.2 moles No. of ions= 0.2 × 6.022 ×1023 ⇒ 1.2042 ×1023 ions