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A) Determine the concentration in molarity (M) of a crystal violet solution based on a measured absorbance of 0.545
at 590 nm at a path length of 1.0 cm .
B) Based on your determined concentration (in molarity,M), how many milligrams of crystal violet were dissolved in the
2 mL of water used to prepare the sample for the cuvette?
C) What is the ppm concentration? The molar extinction coefficienet is 87,000 M–1cm–1 and the
molecular weight is 407.98 g/mol. The density of water is 1.00 g/mL.


  • chemistry -

    A = ebc
    0.545 = 87,000 x 1 x M
    Solve for M = approximately 6E-6 but you need to do it more accurately.

    M = mols/L
    6E-6 = mol/0.002
    mols = about 1.25
    grams = mols x molar mass = about 5E-6 g = about 5E-3 mg.

    5E-6g = 5E-3 mg in 2 mL. Convert that to mg/L to obtain ppm.

  • chemistry -

    THANK YOU!!!!

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