Question

We are testing for the presence of water-soluble “phenolites” (molar mass = 260 g/mole) in mashed potatoes. We know that phenolites have a molar extinction coefficient of 25,000 in a 1 cm light path. We extract 10 g of mashed potatoes in 50 mL of water, and filter the extract to obtain an optically clear solution. We then dilute this solution 1:10 (1 mL of extract + 9 mL water) and read the absorbance on a spectrophotometer (Abs = 0.75). We do not have a standard for phenolites from which to run a standard curve. Calculate the concentration of phenolites in this mashed potatoes and express the data in molar concentration (mmole/kg) and in ppm.

Answer 1

According to beer's-lambert law

A = c l

Where

A = Absorbance = 0.75

= Molar extinction coefficient = 25,000 (unit not given, let it be M^-1 cm^-1)

c = ?

l = path length = 1 cm

So, the concentration is

c = A / l

= (0.75) / (25000 x 1)

= 3.0 x 10^-5 M

So, 3.0 x 10^-5 M is the “phenolites” cocentration in 10 mL solution.

So, in 50 mL of solution, the concentration is expressed as

M1V1 = M2 V2

M2 = M1V1 / V2

= (3.0 x 10^-5 M) (10 mL) / (50 mL)

= 6.0 x 10^-6 M

= 6.0 x 10^-6 mole / Liter

= 6.0 x 10^-6 mole / Kg (density of water = 1)

= 6.0 x 10^-3 mmole / Kg (1 mole = 1000 mmole)

Parts per million: The number of milligrams of solute per kg of solution = 1 ppm.

Now,

Molarity = 3.0 x 10^-5 M

So, 3.0 x 10^-5 moles in 1 liter

Now,

Moles = 3.0 x 10^-5 moles

Molar mass = 260 g/mole

So, Mass = Moles x Molar mass

= (3.0 x 10^-5 moles) x (260 g/mole)

= 0.0078 g

Since, 0.0078 g is present in per Kg (or per liter). So, in ppm it is 0.0078 ppm.