The effective molar mass of ashes has units of grams of ashes per mole of base provided. The mass of the ashes in the experiment is given, 20.4 g. The rest of the work involves determining the amount of base provided, in moles. We will first determine the moles of base used in the titration. What volume of acid was used in the experiment

molar mass of  unknown monoprotic acid = 114.1 g / mole.

Explanation:

Note: This question is incomplete and lacks necessary data to solve for the required calculation. However, I have similar question on the internet and seen the question completely and will be using that data to solve for this question in order to solve required calculation. Besides that, complete question asks us to solve for molar mass of the acid used.

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Calculation:

Volume = 34.81mL

34.81 mL of 0.4346 M potassium hydroxide

= 0.03481 L * 0.4346 mole / L

= 0.01513 mole.

Balance equation:

HA + KOH ----> KA + H2O

As we know from the question,

one mole HA neutralize with 1 mole KOH.

mole of unknown monoprotic acid = 0.01513 mole.

And we know that the formula for mole is:  

mole = mass / molar mass

Making molar mass as the subject:

 Molar mass = mass / mole

Molar mass = 1.726 g / 0.01513 mole

Molar mass = 114.1 g / mole.

So,  

Molar mass of  unknown monoprotic acid = 114.1 g / mole.

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yes it is balanced