A 0.388 g sample of a monoprotic acid is dissolved in water and titrated with 0.280 M NaOH. What is the molar mass of the acid if 10.5 mL of the NaOH solution is required to neutralize the sample? A flask with a solution sits on the base of a ring stand. A buret filled with liquid is suspended above the flask by the ring stand.

The molar mass of the monoprotic acid is 131.97 g/mol

Explanation:

Step 1: Data given

Mass of the sample monoprotic acid = 0.388 grams

Molarity of NaOH = 0.280 M

Volume of NaOH = 10.5 mL = 0.0105 L

Step 2: The balanced equation

HX + NaOH → NaX + H2O

Step 3: Calculate moles NaOH

Moles NaOH = molarity NaOH * volume NaOH

Moles NaOH = 0.280 M * 0.0105 L

Moles NaOH = 0.00294 moles

Step 3: Calculate moles HX

For 1 mol NaOH we need 1 mol HX to react

For 0.00294 moles NaOH we need 0.00294 moles HX to react

Step 4: Calculate molar mass of the acid

Molar mass = mass / moles

Molar mass = 0.388 grams /0.00294 moles

Molar mass = 131.97 g/mol

The molar mass of the monoprotic acid is 131.97 g/mol

the answer would be b i believe