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

The Correct Option is C

(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 C_l2 (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.