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 molar of HClO4=0.803x40x10-3=32.12x10-3 mol
So the final molarity of HClO4=32.12x10-3 mol/660.00x10-3 L=0.0487 mol/L=0.0487 M
The molar of NaSCN=0.351x20x10-3 mol
So the final molarity=0.351x20x10-3/100x10-3=0.351/5=0.0702 M
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