6. How many moles of Sr(NO3)2 would be used in the preparation of 2.50 L of a 30.5 M solution?
Each liter of 30.5 M solution contains 30.5 moles of strontium nitrate, so in 2.50 L there will be
30.5*2.50 = 76.25 mol of Sr(NO3)2.
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