Coagulation value of the electrolytes $$AlC{l_3}$$  and $$NaCl$$  for $$A{s_2}{S_3}$$  sol are 0.093 and 52 respectively. How many times $$AlC{l_3}$$  has greater coagulating power than $$NaCl ?$$

Solution :
$$\eqalign{ & \frac{{{\text{Coagulation power of}}\,AlC{l_3}}}{{{\text{Coagulation power of}}\,NaCl}} \cr & = \frac{{{\text{Coagulation value of}}\,NaCl}}{{{\text{Coagulation value of}}\,AlC{l_3}}} \cr & = \frac{{52}}{{0.093}} = 559.13 \approx 560 \cr} $$