Question

A sample of 0.281 gg of an unknown monoprotic acid was dissolved in 25.0 mLmL of water and titrated with 0.0950 M NaOH NaOH. The acid required 30.0 mLmL of base to reach the equivalence point.What is the molar mass of the acid?

Answer 1

Answer:

98.6 g/mol.

Explanation:

Equation of the reaction

HX + NaOH--> NaX + H2O

Number of moles = molar concentration × volume

= 0.095 × 0.03

= 0.00285 moles

By stoichiometry, 1 mole of HX reacted with 1 mole of NaOH. Therefore, number of moles of HX = 0.00285 moles.

Molar mass = mass ÷ number of moles

= 0.281 ÷ 0.00285

= 98.6 g/mol.