JAMB - Chemistry (1986 - No. 42)
The number of atoms of chlorine present in 5.85g of NaCl is [Avogadro's Number = 6.02 x 10\(^{23}\)]
6.02 x 1022
5.85 x 1023
6.02 x 1023
5.85 x 1024
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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