Question

Determine the amount of solute in the following samples: 500 ml of 0.2 M KI (potassium iodide) 220 ml of 0.03 M H2SO4 (sulfuric acid) 2.5 l of 0.1 M MgNO3 (magnesium nitrate) 1 ml of 5 M HCl (hydrochloric acid)

Answer 1

Amount of KI present in the given sample = (M*V/1000)*F.wt of KI

          = (500*0.2/1000)*166

          = 16.6 g

Amount of H2SO4 present in the given sample = (M*V/1000)*M.wt of H2SO4

= (220*0.03/1000)*98

= 0.65 g

Amount of Mg(NO3)2 present in the given sample = (M*V/1000)*M.wt of Mg(NO3)2

= (2.5*0.1)*148.3

= 37.075 g

Amount of HCl present in the given sample = (M*V/1000)*M.wt of HCl

   = (1*5/1000)*36.5

= 0.1825 g