Answer to Question #134570 in General Chemistry for A

Question

A 6.08×10-2 g sample of an unknown monoprotic acid is dissolved in water and titrated with standardized sodium hydroxide. The equivalence point in the titration is reached after the addition of 31.7 mL of 3.57×10-2 M sodium hydroxide to the sample of the unknown acid. Calculate the molar mass of the acid.

Please Give your answer to three significant figures.

Answer 1

$Molar mass =mass×molarity$

Mass=6.08×10^-2 g 

Molarity =3.57×10^-2 M

Molar mass=(3.57×10-2 M) ×(6.08*10^-2)=0.00217