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Solved: A 0.125-g sample of a monoprotic acid of unknown molar mass is dissolved in

Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo J Tro ISBN: 9780321910295 34

Solution for problem 95P Chapter 14

Introductory Chemistry | 5th Edition

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Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo J Tro

Introductory Chemistry | 5th Edition

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Problem 95P

Problem 95P

A 0.125-g sample of a monoprotic acid of unknown molar mass is dissolved in water and titrated with 0.1003 M NaOH. The equivalence point is reached after adding 20.77 mL of base. What is the molar mass of the unknown acid?

Step-by-Step Solution:
Step 1 of 3

Solution:

Find: we have to calculate the molar mass of unknown acid.

Given:

Mass of monoprotic acid = 0.125 g

Concentration of NaOH = 0.1003 M

Volume = 20.77mL = 0.02077 L

(0.1003 M NaOH)(.02077 L)= .002083  

Then I take the .125 g sample of unknown acid and divide that by .002083

.125/.002083  = 60.0 g/mol of unknown acid

Thus the molar mass of unknown acid is 60.0 g/mol

Step 2 of 3

Chapter 14, Problem 95P is Solved
Step 3 of 3

Textbook: Introductory Chemistry
Edition: 5
Author: Nivaldo J Tro
ISBN: 9780321910295

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Solved: A 0.125-g sample of a monoprotic acid of unknown molar mass is dissolved in