For a chemical reaction $${t_{\frac{1}{2}}}$$  is 2.5 hours at room temperature. How much of the reactant will be left after 7.5 hours if initial weight of reactant was $$160\,g?$$

Solution :
$$\eqalign{ & {\text{Using the relation}} \cr & \left[ A \right] = {\left[ A \right]_0}{\left( {\frac{1}{2}} \right)^n}\left[ {n = {\text{number of half - lives}}} \right] \cr & T = n \times {t_{\frac{1}{2}}} \cr & {\text{Here,}}\,n = \frac{{7.5}}{{2.5}} = 3 \cr & \therefore \,\,\left[ A \right] = 160 \times {\left( {\frac{1}{2}} \right)^3} \cr & = 160 \times \frac{1}{8} \cr & = 20\,g \cr} $$