You took 0.99ml of a hexokinase assay solution and added 10 ul of an unknown glucose sample. The absorbance at 340nm is 0.422 the extinction coefficient is 6.22x10^3 the molecular mass of glucose is 180.16g/mol. What is the concentration of glucose in the original sample...
In moles/liter? And in mg%?
To find the concentration of glucose in the original sample in moles per liter (M), we can use the formula:
Concentration (M) = (Absorbance / Extinction Coefficient) / Path length
Given:
Absorbance = 0.422
Extinction Coefficient = 6.22 x 10^3 M^-1 cm^-1
Path length = 1 cm
First, let's calculate the absorbance:
Absorbance = 0.422
Next, let's calculate the concentration in molar units:
Concentration (M) = (0.422 / (6.22 x 10^3)) / 1
This gives us the concentration of glucose in moles per liter.
Now, to find the concentration in mg%, we need to convert moles to grams and then to mg%.
Molecular mass of glucose = 180.16 g/mol
Concentration (mg%) = (Concentration (M) * Molecular mass * 10^3) / Volume of sample
Given:
Volume of sample = 0.99 ml = 0.99 cm^3
Now, let's calculate the concentration in mg%:
Concentration (mg%) = (Concentration (M) * Molecular mass * 10^3) / (0.99)
This gives us the concentration of glucose in mg%.
By following these steps, you can find the concentration of glucose in the original sample in moles per liter (M) and in mg%.