For an experiment I was doing a UV spec. To start out i had .1116 grams of produ

Question

For an experiment I was doing a UV spec. To start out i had .1116 grams of product, and dissolved it with 5ml of acetone in a 5ml volumetric flask. So my molarity would be at .06406 M. My absorbance was very large (around 8) so i had to dilute it, so in a separate volumetric flask i put 2 ml of solution and 3 ml of acetone and my peak was still around 3 (my target absorbance was below 1). So again i diluted it with 1 ml of that solution and 4 ml of acetone and finally got it in my absorbance range. I am having trouble finding my final molarity though, any help?? thanks.

Explanation / Answer

Initial : 5ml of 0.06406 M

1st dilution:

2 ml of 0.06406 M diluted to 5 mL

Let the final molarity = M2

So,

M1V1 = M2V2

2 mL x 0.06406 M = M2 x 5 mL

M2 = (2 mL x 0.06406 M) / 5 mL = 0.025624 M

2nd dilution:

1 ml of 0.025624 M diluted to 5 mL

Let the molarity after second dilition = M3

So,

M2V2 = M3V3

1 mL x 0.025624 M = M3 x 5 mL

M3 = (1 mL x 0.025624 M) / 5 mL = 0.0051248 M

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