• # question_answer 64)   16 g of oxygen has same number of molecules as in : (a) 16 g of $CO$              (b) 28 g of ${{N}_{2}}$ (c) 14 g of ${{N}_{2}}$                  (d) 1.0 g of ${{H}_{2}}$

(c, d) Number of molecules of ${{O}_{2}}$ in 16 g ${{O}_{2}}$ $=\frac{16}{32}\times 6.023\times {{10}^{23}}$ $=0.5\times 6.023\times {{10}^{23}}$ 0.5 moles are present in 14 g ${{N}_{2}}$ and 1 g ${{H}_{2}}$, hence these samples will also have $(0.5\times 6.023\times {{10}^{23}})$) molecules. (c) Number of molecules of ${{N}_{2}}$                              $=\frac{w}{mol.wt}\times 6.023\times {{10}^{23}}$                 $=\frac{14}{28}\times 6.023\times {{10}^{23}}$                 $=0.5\times 6.023\times {{10}^{23}}$ Number of molecules of ${{H}_{2}}=\frac{1}{2}\times 6.023\times {{10}^{23}}$ $=0.5\times 6.023\times {{10}^{23}}$