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?
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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