Question

The standard solution of 2.50 × 10-5 M of Red-40 dye (molar mass = 496.42 g.mol-1) has an absorbance of 0.467 at 510 nm using 1.00 cm cuvette. Determine the mass percentage of Red-40 dye in a Kool-Aid packet if a solution of an aliquot (0.1245 g) of Kool-Aid powder dissolved in 1 L of water has an absorbance of 0.248 at 510 nm

Answer 1

A =  ε l C
0.467 = ε * 1cm * 2.50*10^-5 mol/L
ε = 18680 L mol^-1 cm^-1
For  Red-40 dye in a Kool-Aid packet,
0.248 = 18680 L mol^-1 cm^-1 * 1cm * C
Concentration, C = 1.3276*10^-5 mol/L or M
It is given that this is dissolved in 1 L solution so,
mol of Red-40 dye = 1.3276*10^-5 mol/L * 1L = 1.3276*10^-5 mol
gms of Red-40 dye = 1.3276*10^-5 mol * 496.42 gm/mol = 6.59058*10^-3 gm
mass %  of Red-40 dye in a Kool-Aid packet = (6.59058*10^-3gm / 0.1245gm)*100
Mass %  of Red-40 dye in a Kool-Aid packet = 5.2936 %