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:

1. 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.
2. 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.
3. 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 $$4$$.

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 $$\Delta n = \Delta c\times V = \Delta c \times \pu{1.2 mL}$$