Answered: Calculate the analytical molar…

World of Chemistry, 3rd edition

3rd Edition

ISBN:9781133109655

Author:Steven S. Zumdahl, Susan L. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan L. Zumdahl, Donald J. DeCoste

Chapter15: Solutions

Section: Chapter Questions

Problem 13A

See similar textbooks

Related questions

Concept explainers

Question

Calculate the analytical molar concentrations of trichloroacetic acid in an aqueous solution that contains 152.9 mg of trichloroacetic acid,Cl₃CCOOH (163.4 g/mol), in 19.3 mL (the acid is 73% ionized in water). Indicate units

Expert Solution

Step 1

Analytical molar concentration of a solution refers to the initial number of moles of a solute in 1 L of the solution. So, it doesn't matter how much the reactant reacts to evaluate the analytical molar concentration.

Step 2

Given:
Mass of solute (chloroacetic acid) in the solution = 152.9 mg = 152.9 X 10^-3 g

Molar mass of solute = 163.4 g/mol

Volume of solution = 19.3 mL = 19.3 X 10^-3 L

Thus, analytical molar concentration of the solution:

$M = \frac{n u m b e r o f m o l e s o f s o l u t e}{V o l u m e o f s o l u t i o n} = \frac{m a s s o f s o l u t e}{m o l a r m a s s o f s o l u t e \times v o l u m e o f s o l u t i o n} = \frac{152.9 \times 10^{- 3} g}{163.4 g / m o l \times 19.3 \times 10^{- 3} L} = 0.048 m o l / L = 0.048 M$

Step by step

Solved in 3 steps

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.

Similar questions