How to calculate amount of nonanoic acid needed to prepare sodium nonanoate salt

I have this procedure for making sodium heptanoate, but I need to adapt it to make sodium nonanoate.


The protocol is as following:


1. Dissolve $\ce{NaOH}$ in pure ethanol
2. After all $\ce{NaOH}$ has dissolved, add acid
3. Precipitates will form, dry the mixture in the oven
4. Dilute the salt to a working solution


Here is the chemical equation I wrote: (please correct me if I am wrong)


for sodium heptanoate preparation:
$$\ce{NaOH + C2H5OH -> C2H5ONa + H2O}$$


$$\ce{C7H14O2 + C2H5ONa -> C7H13NaO2 + C2H5OH}$$


for sodium nonanoate preparation:
$$\ce{NaOH + C2H5OH -> C2H5ONa + H2O}$$


$$\ce{C9H18O2 + C2H5ONa -> C9H17NaO2 + C2H5OH}$$


Question: 


1. How can I calculate the amount (in ml) of nonanoic acid I need to get 1L of solution?
2. Does the amount of $\ce{NaOH}$ dissolved in absolute ethanol matter? Because I have found related protocol using 40 g $\ce{NaOH}$ and some using 60 g $\ce{NaOH}$


Please help me to understand step by step, since I am not a chemistry student. I am a biology student and I need to prepare the salt for my experiments. 


I have included the picture of the protocol for your reference. \*I know the answer is 180ml for heptanoic acid preparation, but **how** do I get to the answer? I need to know the calculation steps, so I can calculate the ml of nonanoic acid I need to make sodium nonanoate.



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> [![protocol](https://i.stack.imgur.com/pnb9C.jpg)](https://i.stack.imgur.com/pnb9C.jpg)
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