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?
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 :)
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