Question

How many grams of potassium iodide (KI) are required to prepare 50 mL of a 0.30 M
solution?

Answer 1

137 g

Step-by-step explanation:

Since1 L=103 mLyou can say that your sample must contain 550.0mL solution⋅1.50 moles KI103mL solution=0.825 moles KITo convert the number of moles of potassium iodide to grams, use the molar mass of the compound.0.825moles KI⋅166.003 g1mole KI=137 g−−−−−