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The mixture contains 62 % one isomer and 38 % the enantiomer.
Let’s say that the mixture contains 62 % of the (<em>R</em>)-isomer.
Then % (<em>S</em>) = 100 % -62 % = 38 %
ee = % (<em>R</em>) - % (<em>S</em>) = 62 % -38 % = 24 %
Answer:

Explanation:
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In this case, given this is an acid-base neutralization and we are considering a diprotic acid, we can write the following mole-mole relationship:

It means that the moles of acid can be computed given the volume and concentration of NaOH:

It means that the approximate molar mass of the acid is:

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