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Suppose a solution was too concentrated for an accurate reading with the spectrometer. The concentrated solution was diluted by placing 1.00mL of the concentrated solution in 4.00 mL of water. The solution was then placed in the spectrophotometer and an absorbance was obtained and after a few calculations the molar concentration was calculated to be 3.5 x 10^-6 M. What was the concentration of the original stock solution before dilution?
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M\(_1\)V\(_1\)=M\(_2\)V\(_2\)
Yes, that's what I originally thought, but I just wasn't sure. Is the answer 1.8 x 10^-5 M?
I don't know, you tell me. You're solving for: M\(_1\).
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