(3 pts) 6) A solution is prepared by mixing 0.45 mL of 5.0 x 10ⓇM proflavin,...

Question

Question

science chemistry

Answer 1

Answer #1

Number of moles of proflavin present in the mixture

=Molarity x Volume (L)

=5.0 x 10^-6 M x 0.45 mL/1000 mL/L (1 L=1000 mL)

=2.25 x 10^-9 mol

Total volume of solution=0.45 mL + 0.45 mL=0.90 mL

=0.90 mL/1000 mL/L=0.90 x 10^-3 L

Concentration of Proflavin in the mixture

=number of moles/volume of solution (L)

=2.25 x 10^-9 mol/0.90 x 10^-3 L=2.5 x 10^-6 M

So the answer is 2.5 x 10^-6 M

Answer 2

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