If 38.2 ml of 0.248M aluminum sulfate solution is diluted with deionized water to a total...
If 38.2 ml of 0.248M aluminum sulfate solution is diluted with deionized water to a total volume of 0.639L; how many grams of aluminum ion are present in the diluted solution?
Homework Answers
intial final
M1 = 0.248M M2 =
V1 = 38.2ml V2 = 0.639L = 639ml
M1V1 = M2V2
M2 = M1V1/V2
= 0.248*38.2/639 = 0.0148M
no of moles of Al2(SO4)3 = molarity * volume in L
= 0.0148*0.639 = 0.009457moles
mass of Al2(SO4)3 = no of moles * gram molar mass
= 0.009457*342 = 3.23g >>>>answer
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