calculate the total no of ions present in 117g  of   NaCl

1. 1 mole of NaCl has 2 moles of ions, 1 mole of Na⁺ and 1 mole of Cl^-.

2. Number of moles of molecules in 117 g of NaCl = $\frac{117}{58.44}\approx 2 moles$
3. Number of moles of ions = 2(2) = 4 moles of ions.
  
      1 mole = 6.023 $\times$10²³

     So, total number of ions = 4(​ 6.023 $\times$10²³)
                                              = 2.41 ​$\times$10²⁴ ions.