What is the final molarity when 40.00 mL of a 0.803-M HClO_{4} 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?

This questions is simple. It is based off the dilution equation:

CiVi = CfVf

Ci = initial concentration

Vi = initial volume

Cf = final concentration

Vf = final volume.

To find the final concentration/molarity rearrange the equation like so (Cf= (CiVi)/Vf)

Now plug in your values and you should get your answer. For example, in your first question, you should plug in the following numbers

Ci = 0.803M

Vi = 40.00mL

Vf = 660.00mL

The equation should look like so: Cf = (0.803M x 40.00mL)/ 660.00mL

For the second question, it should look like so:

Ci = 0.351M

Vi = 20.00mL

Vf = 100.00mL

The equation should look like so: (Cf = (0.351M x 20.00mL)/ 100.00)

Make sure you always make your units are the same and they cancel out! Cheers!

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