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

vanderwaal's forces of attraction are the electrostatic intermolecular forces that exist in a molecule due to the temporary or permanent dipoles that are developed in a molecule. in , the major intermolecular forces are dipole-dipole interactions as the molecule is polar (bent structure). so the vanderwaal's forces of interaction in between molecules are the permanent dipole-dipole interactions that develop due to the dipole that exists in between the polar h - s bond. the positively charged hydrogen atom is attracted towards the negative end of the dipole that is sulfur of the adjacent molecule. therefore, these forces of attraction are relatively stronger than that of london dispersion forces that exist in non polar molecules.