Question #6563

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

Expert's answer

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

moles

n=m/MW

m=n*MW=0.00625*80=0.5 g

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