Answer :
Answer:
0.05 M is the concentration of iodide in the solution in the test tube.
Explanation:
Molarity of KI = 0.20 M
Volume of KI solution = 2.00 mL = 0.002 L ( 1 mL = 0.001 L)
Moles of KI = n
[tex]Molarity=\frac{\text{Moles of solute }}{\text{Volume of solution in L}}[/tex]
[tex]0.20 M=\frac{n}{0.002 L}[/tex]
[tex]n=0.20 M\times 0.002 mL =0.0004 mol[/tex]
Volume of solutions formed by mixing all the solutions in a test tube = V
V = 2.00 mL+1.00 mL+0.50 mL+0.50 mL+2.00 mL+2.00 mL = 8.00 ml
V = 8.00 mL = 0.008 L
Concentration of KI in the test tube = [KI]
[tex][KI]=\frac{0.0004 mol}{0.008 L}=0.05 M[/tex]
[tex]KI(aq)\rightarrow K^+(aq)+I^-(aq)[/tex]
[tex][I^-]=[KI]=0.05 M[/tex]
0.05 M is the concentration of iodide in the solution in the test tube.