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The number of atoms of chlorine present in 5.85g of NaCl is [Avogadro's Number =...

Chemistry
JAMB 1986

The number of atoms of chlorine present in 5.85g of NaCl is [Avogadro's Number = 6.02 x 10\(^{23}\)]

  • A. 6.02 x 1022
  • B. 5.85 x 1023
  • C. 6.02 x 1023
  • D. 5.85 x 1024
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Correct Answer: Option A
Explanation

                                            NaCl(s) ⇌ Na⁺(aq) + Cl⁻(aq).

Number of moles = \(\frac{Number of atom}{Avogadro number}\)

                          n = \(\frac{N}{L}\)

Recall that Number of moles, n = \(\frac{mass}{Molar mass}\)

Molar mass of NaCl = 23 + 35.5 = 58.5g/mol

                                              n =  \(\frac{5.85}{58.5}\)

                                              n = 0.1 mole

⇒                        N = n x L = 0.1 x 6.02 x 10\(^{23}\)

                           N = 6.02 x 10\(^{22}\)


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