• # question_answer36) Which of the following has largest number of atoms? (i) 1 g of Au (ii) 1 g of Na (iii) 1 g of K (iv) 1 g of $C{{l}_{2}}$

(i) Number of atoms = $\frac{W}{{{A}_{w}}}\times 6.023\times {{10}^{23}}$ $=\frac{1}{197}\times 6.023\times {{10}^{23}}$ $=3.06\times {{10}^{21}}atoms$ (ii) Number of atoms = $\frac{1}{23}\times 6.023\times {{10}^{23}}$ $=2.62\times {{10}^{22}}atoms$ (iii) Number of atoms = $\frac{1}{39}\times 6.023\times {{10}^{23}}$ (iv) Number of molecules = $\frac{W}{{{M}_{w}}}\times 6.023\times {{10}^{23}}$ $=\frac{1}{71}\times 6.023\times {{10}^{23}}$ Number of atoms = $2\times \frac{1}{71}\times 6.023\times {{10}^{23}}$ $=1.67\times {{10}^{22}}atoms$ atoms $\therefore$ 1 g Na has maximum number of atoms.