I dissolved 4g of solid NaOH in 200mL of water. I converted grams of NaOH to moles by 4g/40g/moles= 0.1moles.

I calculated the concentration of NaOH by 0.1/(200mL/1000L)= 0.5 moles/L.
I now have to record the expected mass of benzoic acid required to react in stoichiometric proportion with 20mL of the NaOH solution. Not sure about this part.

To calculate the expected mass of benzoic acid required to react with the NaOH solution, you need to consider the stoichiometry of the reaction between NaOH and benzoic acid.

First, let's determine the balanced chemical equation for the reaction between NaOH and benzoic acid:

NaOH + C6H5COOH → NaC6H5COO + H2O

From the balanced equation, we can see that the mole ratio between NaOH and benzoic acid is 1:1. This means that for every 1 mole of NaOH, we need 1 mole of benzoic acid.

Given that the concentration of the NaOH solution is 0.5 moles/L, and you are using 20 mL of the solution, we first need to convert the volume to liters:

20 mL = 20/1000 L = 0.02 L

Next, we can calculate the number of moles of NaOH in the 20 mL solution:

0.5 moles/L * 0.02 L = 0.01 moles of NaOH

Since the mole ratio between NaOH and benzoic acid is 1:1, the number of moles of benzoic acid required will also be 0.01 moles.

Finally, to find the expected mass of benzoic acid, you need to use the molar mass of benzoic acid, which is 122.12 g/mol:

Mass of benzoic acid = Number of moles * Molar mass
Mass of benzoic acid = 0.01 moles * 122.12 g/mol = 1.2212 g

Therefore, the expected mass of benzoic acid required to react in stoichiometric proportion with 20 mL of the NaOH solution is approximately 1.22 grams.

First, you write the equation.

PhCOOH + NaOH ==> PhCOONa + H2O

So you used 20 mL of 0.5 mole/L which means you used 0.50 x 0.02 = 0.01 mole NaOH.
Since the reaction is a 1:1 mole ratio, it will require 0.01 mole benzoic acid. How many grams is 0.01 mole benzoic acid? You know benzoic acid is about 122 g/mol so 0.01 moles would be about 1.22 g. You can do it exactly.