Answer to Question #116746 in General Chemistry for Aaron

Question

Calculate how many grams of KO2LAiD are needed to make 43.0 mL of a 0.25 M solution.

Answer 1

1)

$$n(KAl(OH)_2) = Vc = 0.25M*0.043L = 0.01075 mol$$

2)

$$m(KAl(OH)_2) = 0.01075mol*93.06138g/mol = 1.00041 g$$