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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? 

Dec 11th, 2014

Seems like you have two questions there??

Regardless, both of them would be answered based on this formula: M1V1=M2V2

First Problem : 40.00 mL x 0.803 M = 660.00 mL x M2 (M2 is the final molarity you are looking for)

Thus, M2 = 40.00 mL x 0.803 M / 660.00 mL = 0.049 M

Second Problem: here, you are looking for the initial concentration which is denoted M1, thus:

M1 x 100.00 mL = 20.00 mL x 0.351 M

M1 = 0.070 M 

Hope the answer was within 5 minutes :)

Good luck!

Dec 17th, 2014

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