The molecular weight of methylene blue is 320 g/mol. The concentration of the or

Question

The molecular weight of methylene blue is 320 g/mol. The concentration of the original sample was 0.05 mg/mol. Deteremine the molarity (M) of your original sample and each dilution.

Original:
Dilution 1 (1:2   3mL of methylene blue + 3mL of water) :
Dilution 2 (1:4 3mL of of 1:2 solution + 3mL of water) :
Dilution 3 (1:8  3mL of 1:4 solution + 3 mL of water) :
Diltuion 4 (1:10 3Ml of original + 4.5 mL of water) :

PLEASE HELP ME & SHOW WORK TO DETERMINE MOLARITY AND I"LL AWARD YOU FULL POINTS!!  

Explanation / Answer

Original:

Volume = 1 mL = 0.001 L

Mass of methylene blue = 0.05 mg = 0.05/1000 = 5 x 10^(-5) mol

Moles of methylene blue = mass/molar mass of methylene blue = 5 x 10^(-5)/320 = 1.56 x 10^(-7) mol


Molarity = moles/volume

= 1.56 x 10^(-7)/0.001 = 1.56 x 10^(-4) M


Dilution 1:

M1 = 1.56 x 10^(-4) M, M2 = ?

V1 = 3 mL, V2 = 3 + 3 = 6 mL


M1V1 = M2V2

1.56 x 10^(-4) x 3 = M2 x 6

Molarity M2 = 7.8 x 10^(-5) M


Dilution 2:

M1 = 7.8 x 10^(-5) M, M2 = ?

V1 = 3 mL, V2 = 3 + 3 = 6 mL


M1V1 = M2V2

7.8 x 10^(-5) x 3 = M2 x 6

Molarity M2 = 3.9 x 10^(-5) M


Dilution 3:

M1 = 3.9 x 10^(-5) M, M2 = ?

V1 = 3 mL, V2 = 3 + 3 = 6 mL


M1V1 = M2V2

3.9 x 10^(-5) x 3 = M2 x 6

Molarity M2 = 1.95 x 10^(-5) M


Dilution 4:

M1 = 1.56 x 10^(-4) M, M2 = ?

V1 = 3 mL, V2 = 3 + 4.5 = 7.5 mL


M1V1 = M2V2

1.56 x 10^(-4) x 3 = M2 x 7.5

Molarity M2 = 6.25 x 10^(-5) M

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