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  • Compute the number of ions present in 5.85 g of sodium chloride.

Compute the number of ions present in 5.85 g of sodium chloride.

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Answer:

5.85g of NaCl=\frac{5.85}{58.5}=0.1moles
Or, 0.1 moles of NaCl particles.
Each NaCl particle is equivalent to one Na^{+} one Cl^{-} i.e. 2 ions.
Total moles of ions = 0.1 \times 2 = 0.2 moles
No. of ions = 0.2 \times 6.022 \times 10^{23} = 1.2042 \times 10^{23} ions

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