Question

You have 3 test tubes. 7.5 ml H2O in tube A, 5.0 ml H2O in tube B, 2.5 ml H2O in tube C. You have a stock solution of....KI (Potassium Iodide) with a concentration 12 mol/L. You place 2.5 mil of the stock solution into tube A.

Question 1: What iscthe concentration of solution A?

You then take 5 ml of solution A (in tube A) and transfer it to tube B.

You mean 2.5 mL ?

You then take 7.5 ml of solution B (in tube B) and tranfet it to tube C.

Question 3: What is the comcrntration of solution C?

Answer 1

1. Test tube A contains 7.5ml H₂O and to it 2.5ml of 12mol/L KI solution is added.

Molarity of KI stock solution= 12mol/L

Therefore, 1L(1000ml) of KI solution contains =12 moles of KI

2.5 mL of KI solution will contain= (12moles/1000mL)x 2.5mL=0.03 moles of KI

Volume of solution in test tube A= (7.5+2.5) mL=10 mL =0.01L

Molarity of KI solution in test tube A= (No. of moles of KI)/(Volume of solution in test tube A in Litres)

=0.03 moles/0.01 L

=3 mol/L

2. Test tube B contains 5 ml H₂O and to it 5ml of 3 mol/L KI solution is added from test tube A.

Molarity of KI solution in test tube A = 3 mol/L

Therefore, 1L(1000ml) of KI solution contains = 3 moles of KI

5 mL of KI solution will contain= (3moles/1000mL)x 5mL=0.015 moles of KI

Volume of solution in test tube B= (5+5 )mL=10 mL =0.01 L

Molarity of KI solution in test tube B= (No. of moles of KI)/(Volume of solution in test tube B in Litres)

=0.015 moles/0.01 L

=1.5 mol/L

3.Test tube C contains 2.5 ml H₂O and to it 7.5ml of 1.5 mol/L KI solution is added from test tube B.

Molarity of KI solution in test tube B = 1.5 mol/L

Therefore, 1L(1000ml) of KI solution contains = 1.5 moles of KI

7.5 mL of KI solution will contain= (1.5moles/1000mL)x7.5mL=0.01125 moles of KI

Volume of solution in test tube C= (2.5+7.5 )mL=10 mL =0.01 L

Molarity of KI solution in test tube C= (No. of moles of KI)/(Volume of solution in test tube C in Litres)

=0.01125 moles/0.01 L

=1.125 mol/L