A0.229-g sample of an unknown monoprotic acid is titrated with 0.112 m naoh. the resulting titration curve is shown here. part a determine the molar mass of the acid. express your answer using two significant figures.

81.8 g/mol

Explanation:

We’re being asked to calculate the molar mass of an unknown acid based on our interpretation of the titration curve.  

Recall that at the equivalence point of a titration:

We can use this equation to calculate the moles of the unknown acid. Since the unknown acid only has a single Ka value, it is a monoprotic acid, which means there will be one equivalence point for the titration.

Mole-to-mole comparison: A mole-to-mole comparison will then show that 1 mole of the unknown acid reacts with 1 mole of Na(OH).

Molarity of NaOH: The molarity of NaOH (the base) is given as 0.112 M or 0.112 mol/L NaOH.

As for the volume of NaOH (the base) added, we can interpret this from the graph:  

we can see that at the equivalence point, 25 mL of the vase was added.

Vbase is more or less 25.00 mL

Recall:  

1 mL = 10-3 L  

Solving for MMacid:  

Again, at the equivalence point of a titration:

Molar mass is defined as the mass in grams of one mole of a substance.  

The units of molar mass are grams per mole, abbreviated as g/mol and uses the formula below:

molar mass acid = 81.8 g/mol

The molar mass of the unknown acid is 81.8 g/mol.

catya that's not even a b or c

water and alcohols have similar properties because water molecules contain hydroxyl groups that can form hydrogen bonds with other water molecules and with alcohol molecules, and likewise alcohol molecules can form hydrogen bonds with other alcohol molecules as well as with water.