Question

Serial dilution is an essential chemistry skill. Use the description given to calculate the concentration of each new solution. Report all answers to three significant figures, with units of M.

The concentration of the initial stock solution is 3.00 M. The first dilution is made by measuring 10.00 mL with a volumetric pipet and diluting in a 100 mL volumetric flask. What is the concentration of the first solution?

The second solution is made by measuring 25.00 mL of the first solution and diluting in a 100 mL volumetric flask. What is the concentration of the second solution?

The third solution is made by measuring 50.00 mL of the second solution and diluting in a 250 mL volumetric flask. What is the concentration of the third solution?

The fourth solution is made by measuring 1.00 mL of the third solution and diluting in a 500 mL volumetric flask. What is the concentration of the fourth solution? (Count the zeros carefully when entering an answer!)

Answer 1

Mole of 3M in 10ml =3*10/1000= 0.03 moles

this must be equal to =M1V1, where M1= concentration after first dilution and V1= volume =100ml= 0.1L

M1= 0.03/0.1= 0.3M

In the second dilution, number of moles = 0.3*25/1000= 0.0075

again M2= 0.0075/0.1 ( V2=100ml=0.1), M2= 0.075M

In the third dilution, number of moles =0.075*50/1000= 0.00375 moles

again M3= 0.00375/0.25 (250ml =0.25L)= 0.015 M

mole in the fouth dilution= 0.015*1/1000=0.000015

Concentration after fourth dilution= 0.000015/0.5 (500ml =0.5 L)=0.00003 M