Question

calculate the number of molecules present in a drop of chloroform chcl3 waiting 0.023 9 gram​

Answer 1

molecuar mass of CHCl3 = 12 + 1 +3 x 35.5 = 119.5g

119.5g contain = 6.022 x 10^23 molecules

thereforce, 0.0239g contain = 6.022 x 10^23 x 0.0239 / 119.5

                                                  = 12.044 x 10^19 molecules