A) Determine the concentration in molarity (M) of a crystal violet solution based on a measured absorbance of 0.545

Question

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.

PLEASE HELP!!!! :)

Answer 1

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

b.
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.

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

Answer 2

THANK YOU!!!!