Question

How many grams of berylium chloride are needed to make 125 mL of a .050 M solution? And how do I get the answer?

Answer 1

0.05М = 0.05 moles in 1000 ml,so in 125 ml will be 125*0.05/1000=0.00625
moles
n=m/MW
m=n*MW=0.00625*80=0.5 g