Answer:
the expected absorbance of the solution = 0.17
Explanation:
From the information given:
Using Beer's Lambert Law, we have
A = ∈CL
where;
A = Absorbance
∈ = extinction coefficient
C = concentration
L = cell length
Since Absorbance is associated with concentration.
Assuming the measurement were carried out in the same solution; Then ∈ and L will be constant and A ∝ C ----- (1)
Let consider the concentration to be C (mol/L)
5.0 mL of a Sports Drink = 5.0 mL × C (mol)/1000 mL
= 5C/1000 mL
was diluted with water to 10.0 mL
So, when diluted with water to 10.0 mL; we have:
The new concentration to be : ![\dfrac{(5 C \times 1000) \ mol }{(1000 \times 10 \times 1000)\ mL}](https://tex.z-dn.net/?f=%5Cdfrac%7B%285%20C%20%5Ctimes%201000%29%20%5C%20mol%20%7D%7B%281000%20%5Ctimes%2010%20%5Ctimes%201000%29%5C%20%20mL%7D)
Since :1000mL = 1 L
The new concentration = ![\dfrac{C \ mol }{2 \ L}](https://tex.z-dn.net/?f=%5Cdfrac%7BC%20%5C%20mol%20%7D%7B2%20%5C%20L%7D)
As stated that the initial absorbance reading
= 0.34
The expected absorbance reading will be
= ???
From (1)
A ∝ C
∴
![\dfrac{A_2}{A_1}=\dfrac{C_2}{C}](https://tex.z-dn.net/?f=%5Cdfrac%7BA_2%7D%7BA_1%7D%3D%5Cdfrac%7BC_2%7D%7BC%7D)
![A_2 = \dfrac{A_1}{C}](https://tex.z-dn.net/?f=A_2%20%3D%20%5Cdfrac%7BA_1%7D%7BC%7D)
![A_2 = \dfrac{0.34}{2}](https://tex.z-dn.net/?f=A_2%20%3D%20%5Cdfrac%7B0.34%7D%7B2%7D)
![A_2 = 0.17](https://tex.z-dn.net/?f=A_2%20%3D%200.17)
Thus ; the expected absorbance of the solution = 0.17