Which of the following has largest number of atoms? 1. 1g of Au 2. 1g of Na 3. 1g of Li 4. 1g Cl2
Asked by Prachi Panwar | 23rd May, 2013, 07:36: PM
Expert Answer:
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
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
Answered by | 24th May, 2013, 10:37: AM
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