Which of the following has largest number of atoms? 1. 1g of Au 2. 1g of Na 3. 1g of Li 4. 1g Cl2

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 × 10²³ atoms (Avogadro's Number) 

Thus, 0.005 mole Au is (0.005 mol x 6.022 × 10²³ atoms/mol) = 3.057 x 10²¹ 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 X 10²³  = 2.4 x 10²² atoms

 

3. MW of Li = 6.94

 

Moles = 1/ 6.94 moles = 0.144

 

No of atoms = 0.144 x  6.022 X 10²³   = 8.6 x 10²² atoms

 

4. MW of Cl₂ = 71 g/moles

 

Moles = 1 / 71 = 0.0140

No of molecules of Cl₂ = 0.0140 x 6.022 X 10²³ = 8.4 x 10²¹

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

 

No of atoms  = 2 x 8.4 x 10²¹ = 16.8 x 10²¹ = 1.68 x 10²²

 

 

So, largest no of atoms are in 1 g Li