Question

Which one of the following will have the largest number of atoms?

• A. 1 g Au (s)
• B. 1 g Na (s)
• C. 1 g Li (s)
• D. 1 g of Cl2 (g)

Answer 1

Answer: (C)

1 g Li (s)

Answer 2

(i) 1 g of Au (s) = $\frac {1}{197}$ mol of Au (s)

= $\frac {6.022 × 10^{23}}{197}$ atoms of Au (s)
= 3.06 × 1021 atoms of Au (s)

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(ii) 1 g of Na (s) = $\frac {1}{23}$ mol of Na (s)

= $\frac {6.022 × 10^{23}}{323}$ atoms of Na (s)
= 0.262 × 1023 atoms of Na (s)
= 26.2 × 1021 atoms of Na (s)

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(iii) 1 g of Li (s) = $\frac 17$ mol of Li (s)

= $\frac {6.022 × 10^{23}}{7}$ atoms of Li (s)
= 0.86 × 1023 atoms of Li (s)
= 86.0 × 1021 atoms of Li (s)

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(iv) 1 g of Cl2 (g) = $\frac {1}{71}$ mol of Cl2 (g)
(Molar mass of Cl2 molecule = 35.5 × 2 = 71 g mol-1)

= $\frac {6.022 × 10^{23}}{71}$ molecules of Cl2 (g)
= 0.0848 × 1023 molecules of Cl2 (g)
= 8.48 × 1021 molecules of Cl2 (g)
As one molecule of Cl2 contains two atoms of Cl.
Number of atoms of Cl = 2× 8.48 × 1021 =16.96 × 1021 atoms of Cl

Hence, 1 g of Li (s) will have the largest number of atoms.