Question

The molecular weight of methylene blue is 320 g/mol. The concentration of the original sample was 0.005 mg/ml in 10 ml. Determine the molarity (M) of your original sample and each dilution.

Original:
Dilution 1 (1:2 add 3mL of methylene blue + 3mL of water) :
Dilution 2 (1:4 use 3mL of of 1:2 solution + 3mL of water) :
Dilution 3 (1:8 use 3mL of 1:4 solution + 3 mL of water) :
Dilution 4 (1:10 use 0.5 ml of the original methylene blue + 4.5 mL of water) :

Please show all work!!

Answer 1

Molarity= number of moles/volume of solution in litres

number of moles= mass in gram/gram molecular mass

gram molecular mass= 320 g/mol.

mass = concentration of original solution volume of the solution

volume of the solution= 10 mL

concentration of the origional solution = 0.005 mg/ml

mass of solute=0.005 mg/ml 10

= 0.05 mg

1 mg= 10^-3 g

mass of solute= 0.05 10^-3 g

number of moles=  0.05 10^-3 g/320 g/mol.

= 1.5625 10^-7

1 mL= 10^-3 L

10 mL= 10 10^-3

Molarity of original solution= 1.5625 10^-7 / 1010^-3

= 1.5625 10^-5 M

molarity of diluted samples= molarity of original solution dilution of the solution

for dilution 1

dilution=1/2

molarity= molarity of original solution dilution of the solution

Molarity=1.5625 10^-5 M 1/2

= 7.8125 10^-6 M

for dilution 2

dilution=1/4

molarity= molarity of original solution dilution of the solution

Molarity=1.5625 10^-5 M 1/4

= 3.90625 10^-6 M

for dilution 3

dilution=1/8

molarity= molarity of original solution dilution of the solution

Molarity=1.5625 10^-5 M 1/8

= 1.953125 10^-6 M

for dilution 4

dilution=1/10

molarity= molarity of original solution dilution of the solution

Molarity=1.5625 10^-5 M 1/10

= 1.562510^-6 M