Question

The concentration of dye in Solution A is 20.926 M. A serial dilution is performed to make Solutions B and C. In the first dilution, 7 mL of Solution A is diluted with 12 mL water to make Solution B. Then, 2 mL of Solution B is then diluted with 1 mL of water to make Solution C. What is the concentration of dye in Solution C? Provide your response to three significant figures in units of molarity.

Answer 1

Molarity of solution A=20.926 M

Volume of solution A taken=7 mL=7 mL/1000 mL/L=0.007 L

(1 L=1000 mL)

Number of moles of dye in 7 mL of solution A=Molarity x Volume of solution (L)

=20.926 M x 0.007 L=0.146 mol

Volume of solution B=Volume of solution A+ 12 mL

=7 mL+12 mL=19 mL

=19 mL/1000 mL/L=0.019 L

Molarity of solution B=number of moles of dye/Volume of solution (L)

=0.146 mol/0.019 L=7.684 M

Number of moles of dye in 2 mL of solution B (with molarity 7.684 M)

=Molarity x Volume of solution (L)

=7.684 M x 2 mL/1000 mL/L=0.015 mol

Solution C is prepared by adding 1 mL water to 2 mL of solution B

So total volume of solution C=3 mL=3 mL/1000 mL/L=0.003 L

Molarity of solution C=number of moles of dye/volume of the solution (L)

=0.015 mol/0.003 L=5.00 M (answer in three significant figures)

So concentration of dye in solution C=5.00 M