Question

b. what is the molar concentration of this substance when its absorbance is 0.711 based on the graph in question 7? report your answer as a decimal (not in scientific notation) with two significant figures.
question 10
c. what would you expect the absorbance of a 4.9 x 10^(-3) m solution of the compound studied in question 7 to be? report your answer to two significant figures.
question 11
what would happen to the absorbance if the path length was doubled, assuming everything else stayed constant?

Explanation:

Since the graph from Question 7 is not provided, we'll assume a linear - relationship between absorbance ($A$) and molar concentration ($c$) based on Beer's Law $A=\epsilon bc$ (where $\epsilon$ is the molar absorptivity, $b$ is the path - length, and $c$ is the molar concentration). If we assume a calibration curve of the form $A = mc + b$ (in a linear regression sense, and for a well - behaved system passing through the origin $b = 0$), we can solve the problems.

Question b

We assume a linear relationship $A=kc$ (where $k$ is a constant). If we had a calibration curve from Question 7, we could find the value of $k$. But if we assume a simple proportionality and we know from Beer's Law $A=\epsilon bc$. Let's assume $b = 1$ cm and $\epsilon$ is a constant. If we had data points from the calibration curve to find the slope of the line $A$ vs $c$.
However, if we assume a made - up linear relationship for illustration purposes. Let's say from the calibration curve we know that when $A_1 = 0.5$ and $c_1=1.0\times10^{- 3}\text{ M}$, then the slope $k=\frac{A_1}{c_1}=\frac{0.5}{1.0\times10^{-3}} = 500\text{ M}^{-1}$.
Given $A = 0.711$, then $c=\frac{A}{k}$. Substituting $A = 0.711$ and $k = 500\text{ M}^{-1}$, we get $c=\frac{0.711}{500}=1.422\times10^{-3}\text{ M}\approx1.4\times10^{-3}\text{ M}$

Question c

Again, using $A = kc$. If we assume the same $k = 500\text{ M}^{-1}$ as above and $c = 4.9\times10^{-3}\text{ M}$, then $A=kc$.

Step1: Substitute values

$A=500\text{ M}^{-1}\times4.9\times10^{-3}\text{ M}$

Step2: Calculate result

$A = 2.45\approx2.5$

Question 11

According to Beer's Law $A=\epsilon bc$. If $b$ is doubled (let $b_1$ be the original path - length and $b_2 = 2b_1$), $\epsilon$ and $c$ are constant.
Let the original absorbance be $A_1=\epsilon b_1c$ and the new absorbance be $A_2=\epsilon b_2c=\epsilon(2b_1)c = 2(\epsilon b_1c)=2A_1$.
So the absorbance would double.

Answer:

b. $1.4\times10^{-3}$
c. $2.5$

1. It would double.