The molar extinction coefficient (ε) of bilirubin at 453 nm is 60 M^-1 cm^-1. This means that if a solution of bilirubin with a concentration of 1 M has an optical density (OD) of 1 at 453 nm, the absorbance of a solution with a concentration of 1 M will be 60 at 453 nm.
Now, let's use the absorbance value at 700 nm to calculate the concentration of bilirubin:
A = εbc
where A is the absorbance, ε is the molar extinction coefficient, b is the path length in cm, and c is the concentration in M.
We know that at 700 nm, A = 0.5, ε = unknown, b = unknown, and we want to find c.
To solve for c, we need to know the path length (b) at 700 nm, which is not given in the question. Assuming a path length of 1 cm, we can rearrange the equation as:
c = A/εb
c = 0.5/εb
We can use the molar extinction coefficient of bilirubin at 453 nm to estimate its molar extinction coefficient at 700 nm using the following equation:
ε2 = ε1 x (λ2/λ1)^p
where ε1 is the molar extinction coefficient at λ1, ε2 is the molar extinction coefficient at λ2, and p is the spectral exponent, which is typically between 1 and 2 for biological molecules.
Assuming p = 2, we can calculate ε2 as:
ε2 = 60 x (700/453)^2
ε2 = 402 M^-1 cm^-1
Substituting this value and assuming a path length of 1 cm, we get:
c = 0.5/402x1
c = 0.00124 M
Therefore, the molarity of bilirubin is 0.00124 M.