You need to make an aqueous solution of 0.192 M barium sulfide for an experiment in lab, using a 500 mL volumetric flask. How mu
ch solid barium sulfide should you add?
1 answer:
Answer:
Explanation:
Molecular weight of barium sulphide = 169
500 mL of .192 M barium sulfide = .5 x .192 moles of barium sulphide
= .096 moles of barium sulfide
= .096 x 169 gram of barium sulfide
= 16.22 grams of barium sulfide .
We shall have to add 16.22 gram .
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