II. Serial Ex: (a) A stock solution of 3.65 x 10 M salicylic acid is diluted by transferring 10.0 mL of the stock solution into a beaker and adding 40.0 mL of water. The beaker is labeled dilution 1 5.00 ml of the solution in the beaker labeled dilution 1 is transferred to another beaker and 15.0 mL of water is added. The beaker is labeled dilution 2. 5.00 mL of the solution in the beaker labeled dilution 2 is transferred to another beaker and 15.0 mL of water is added. The beaker is labeled dilution 3. 1.00 mL of the solution in the beaker labeled dilution 3 is transferred to another beaker called dilution 4 and enough water is added so that the final volume is 5.00 mL (b) (c) (d) What are the concentrations of each diluted soln? Solution a) dilution 1 M1V1 = M2V2 M1 = molarity of stock solution = 3.65*10^-4 M V1 = volume of stock solution taken = 10ml M2 = molarity of dilution 1 = ? V2 = final volume of solution= 40 +10 = 50 ml (3.65*10^-4*10) = (M2*50) M2 = 7.3*10^-5 M b) dilution 2 M1V1 = M2V2 M1 = molarity of dilution 1 = 7.3*10^-5 M V1 = volume of stock solution taken = 5 ml M2 = molarity of dilution 2 = ? V2 = final volume of solution= 15+5 = 20 ml (7.3*10^-5*5) = (M2*20) M2 = 1.825*10^-5 c) dilution 3 M1V1 = M2V2 M1 = molarity of dilution 2 = 1.825*10^-5 M V1 = volume of stock solution taken = 5 ml M2 = molarity of dilution 2 = ? V2 = final volume of solution= 15+5 = 20 ml (1.825*10^-5*5) = (M2*20) M2 = 4.562*10^-6 d) dilution 4 M1V1 = M2V2 M1 = molarity of dilution 2 = 4.562*10^-6 M V1 = volume of stock solution taken = 1 ml M2 = molarity of dilution 2 = ? V2 = final volume of solution= 5 ml (4.562*10^-6*1) = (M2*5) M2 = 9.124*10^-7 .

1. II. Serial Ex: (a) A stock solution of 3.65 x 10 M salicylic acid is diluted by transferring 10.0 mL
of the stock solution into a beaker and adding 40.0 mL of water. The beaker is labeled dilution 1
5.00 ml of the solution in the beaker labeled dilution 1 is transferred to another beaker and 15.0
mL of water is added. The beaker is labeled dilution 2. 5.00 mL of the solution in the beaker
labeled dilution 2 is transferred to another beaker and 15.0 mL of water is added. The beaker is
labeled dilution 3. 1.00 mL of the solution in the beaker labeled dilution 3 is transferred to
another beaker called dilution 4 and enough water is added so that the final volume is 5.00 mL
(b) (c) (d) What are the concentrations of each diluted soln?
Solution
a) dilution 1
M1V1 = M2V2
M1 = molarity of stock solution = 3.65*10^-4 M
V1 = volume of stock solution taken = 10ml
M2 = molarity of dilution 1 = ?
V2 = final volume of solution= 40 +10 = 50 ml
(3.65*10^-4*10) = (M2*50)
M2 = 7.3*10^-5 M
b) dilution 2
M1V1 = M2V2
M1 = molarity of dilution 1 = 7.3*10^-5 M
V1 = volume of stock solution taken = 5 ml
M2 = molarity of dilution 2 = ?

2. V2 = final volume of solution= 15+5 = 20 ml
(7.3*10^-5*5) = (M2*20)
M2 = 1.825*10^-5
c) dilution 3
M1V1 = M2V2
M1 = molarity of dilution 2 = 1.825*10^-5 M
V1 = volume of stock solution taken = 5 ml
M2 = molarity of dilution 2 = ?
V2 = final volume of solution= 15+5 = 20 ml
(1.825*10^-5*5) = (M2*20)
M2 = 4.562*10^-6
d) dilution 4
M1V1 = M2V2
M1 = molarity of dilution 2 = 4.562*10^-6 M
V1 = volume of stock solution taken = 1 ml
M2 = molarity of dilution 2 = ?
V2 = final volume of solution= 5 ml
(4.562*10^-6*1) = (M2*5)
M2 = 9.124*10^-7