What is the final molarity when 40.00 mL of a 0.803-M HClO4 solution is diluted to 660.00 mL?
Calculate the molarity of diluted NaSCN if 20.00 mL of 0.351 M NaSCN is transfered to a 100.00 mL volumetric flask, diluted to the mark and mixed well?
The dilution relation is : C1 x V1 = C2 x V2 where 1 and 2 are initial and final state
So for HClO4 solution (0.803) x (40) = C2 x (660)
Therefore C2 = 0.805 x 40 /660 = 0.049 M HClO4
Similarly for NaSCN solution: C2 = 20 x 0.351 /100 = 0.0702 M NaSCN
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