Question

I am doing a lab report on Beer's Law. Equation is absorbance= molar extinction coefficient x path length x concentration. In lab, I prepared a stock solution of .25M and filled it up to 100ml. I prepared 3 more diluted solutions. Dilution 1 was .2M where I added 8ml of stock solution and 2 ml of water for total volume of 10ml.

Dilution 2 was 6ml of stock solution and added 4 ml water for total volume of 10ml. Dilution 3 was 4 ml of stock solution and added 6ml of water for total volume of 10 ml.

Now, my teacher wants me to calculate the true concentration in Molarity (M) for each of these solutions. How do I do that?

If it helps, i used 5.940 g of NiCl2*6H2O for stock solution and filled it up to 100 ml.

Additional info if needed:

Dilution 1 had 7.8 ml stock sol

Dilution 2 had 5.9 ml stock sol

Dilution 3 had 3.9 ml stock sol

If needed: Absorbance for dilution 1 was .915

Absorbance for dilution 2 was .654

Absorbance for dilution 3 was .416

Cuvette diameter for path length is 1.5cm

Answer 1

NiCl2.6H2O Stock solution = 5.940 g/237.691 g/mol x 0.1 L

                                            = 0.25 M

Dilution,

Dilution #1, molarity of solution = 0.25 M x 8 ml/10 ml = 0.2 M

Dilution #1, molarity of solution = 0.25 M x 6 ml/10 ml = 0.15 M

Dilution #1, molarity of solution = 0.25 M x 4 ml/10 ml = 0.1 M

Concentration from absorbance

concentration = absorbanc/molar absorptivity x path length

So, given molar absorptivity we can easily calculate true molarity of each solution from absorbance values.