Q:

Spectrophotometric assay by pNPP

Here is a series of related questions that I want to ask.
Background
The activity of acid phosphatase is measured by an enzymatic reaction that converts para-nitrophenyl phosphate (pNPP) to para-nitrophenol (pNP), liberating phosphate. The product, pNP, absorbs light whose wavelength is $\pu{400 nm}$ with an absorption coefficient ($\pu{400 nm}$) of $\pu{19000 M-1 cm-1}$ at extremely alkaline pH. Reaction mixture for an acid phosphatase is slightly acidic. Thus, it must be alkalinized for quantification of pNP.
Two enzyme concentrations are to be examined - They are $\ce{1X}$ and $\ce{0.1X}$ (The $\ce{0.1X}$ enzyme is made by mixing $\pu{1 ml}$ of $\ce{1X}$ enzyme in $\pu{9 ml}$ of $\ce{NaCl}$.) Reaction times for each enzyme is $1$, $10$ and $20$ minutes.
Procedure

Protocol for measurement of acid phosphatase activity:

Mix $\pu{0.12 ml}$ of $\pu{0.5 M}$ $\ce{Na}$ acetate buffer (pH $5.6$) and $\pu{0.24 ml}$ of $\pu{5 mM}$ pNPP in a test tube. Start the reaction by adding $\pu{0.24 ml}$ of an enzyme solution.
After the reaction times of $1$, $10$, and $\pu{20 min}$, respectively, stop the reaction by adding $\pu{0.6 ml}$ of $\pu{0.5 M}$ $\ce{NaOH}$. $\ce{NaOH}$ stops the reaction and converts the pNP produced into a yellow-colored (A400-absorbing) form.
After all reactions are stopped, measure A400 of the samples.

Assay of potato acid phosphatase:
\begin{array}{ll}
\pu{0.5 M} \text{Na acetate buffer (pH 5.6)} & \pu{0.12 ml}\\
\pu{5 mM} \ce{pNPP} & \pu{0.24 ml}\\
\text{Enzyme} & \pu{0.24 ml}\\
\pu{0.m M} \ce{NaOH} & \pu{0.5 ml}\\\hline
\text{Sum} & \pu{1.2 ml}
\end{array}

There are some things I am supposed to calculate after I obtain the results.
First I am going to plot a graph of absorbance versus time and find the slope for each of the lines ($\ce{1X}$ and $\ce{0.1X}$).

I have been asked to find the absorbance change/min/1ml of $\ce{1X}$ enzyme.

My answer: Divide the slope obtained by $0.24$ (the amount of enzyme).

I have to convert the absorbance change to concentration change when
$L$ is $\pu{1 cm}$ and e400 of pNP is $\pu{19000 M-1 cm-1}$.

My answer: This can be found out by $A = eCL$.

The next question is to convert the concentration change to a change in the amount of substance of pNP.

I have no clue how to do this.

Finally, I have to calculate total activity (in moles per minute) in $\pu{4 ml}$ of $\ce{1X}$ enzyme solution.

My answer: Multiply the answer I obtain in the above question by $4$.

A:

For the first subquestion:
Using unit analysis, the answer is very quickly found: The unit of the slope $m$, assuming dimensionless values for the absorbance, is 
$$[m] = \frac{[\Delta y]}{[\Delta x]} = \frac{[\Delta A]}{[\Delta t]} = \frac{1}{\text{min}}$$
Now you need a volume in the denominator as well, so dividing by a volume seems like a good idea. Since you need the change in absorption per $\pu{mL}$ of $\ce{1X}$ solution, you divide the slope of the $\ce{1X}$ reaction by $\pu{0.24 mL}$ for the correct value.

For the subquestion with which you seem to have the most trouble: simply apply $c=n/V$.
You will find the amount of substance changed via
$$ \Delta n = \Delta c\times V = \Delta c \times \pu{1.2 mL}$$

