Question

Which one of the following will have largest number of atoms?
1. 1 g Au (s)
2. 1 g Na (s)
3. 1 g Li (s)
4. 1 g of Cl2 (g)

Answer 1

Answer:

Explanation:

1. Molecular weight of Au = 196.96 g/mol  

Thus, 1g /196.96 g/mol = 0.005 mole of Au

1 mole has 6.022 × 10~23 atoms (Avogadro's Number)  

Thus, 0.005 mole Au is (0.005 mol x 6.022 × 10~23 atoms/mol)

= 3.057 x 10~21 atoms

2. MW of Na = 23 g/mole

No of moles of Na =  1/23

So, atoms of Na in moles = 0.04 x 6.022 × 10~23  

= 2.4 x 10~22 atoms

3. MW of Li = 6.94

Moles = 1/ 6.94 moles = 0.144

No of atoms = 0.144 × 6.022 × 10~23  

= 8.6 x 10~22 atoms

4. MW of Cl2 = 71 g/moles

Moles = 1 / 71 = 0.0140

No of molecules of Cl2 = 0.0140 x 6.022 × 10~23

= 8.4 x 10~21

In 1 molecule of chlorine there are 2 atoms, So,

No of atoms  = 2 x 8.4 x 10~21

= 16.8 x 10~21

= 1.68 x 10~22

Therefore, the largest no of atoms are in 1g of Li.