A 0.2386-g sample contained only NaCl and KBr was dissolved in water and required 50.00 mL of 0.04500 M AgNO3 for complete titration of both halides. If we assume there is x mol of NaCl in the original sample, which of the following statements are CORRECT? There are multiple correct answers. MW(NaCl) = 58.44 g/mol. MW(KBr) = 119.002 g/mol. There is x mol of KBr in the original sample. Ox+(0.2386-58.44x)/119.002-50.00*0.04500 There is (0.002250 - x) mol of KBr in the original sample. 58.44x+119.002(0.002250-x)=0.2386

A 0.2386-g sample contained only NaCl and KBr was dissolved in water and required 50.00 mL of 0.04500 M AgNO3 for complete titration of both halides. If we assume there is x mol of NaCl in the original sample, which of the following statements are CORRECT? There are multiple correct answers. MW(NaCl) = 58.44 g/mol. MW(KBr) = 119.002 g/mol. There is x mol of KBr in the original sample. Ox+(0.2386-58.44x)/119.002-50.00*0.04500 There is (0.002250 - x) mol of KBr in the original sample. 58.44x+119.002(0.002250-x)=0.2386

Chemistry

10th Edition

ISBN:9781305957404

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

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

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Concept explainers

Acid Base Titration

An acid-base titration is the method of quantitative analysis for determining the concentration of an acid and base by exactly neutralizing it with a standard solution of base or acid having known concentration.

Question

A 0.2386-g sample contained only NaCl and KBr was dissolved in water and required 50.00
mL of 0.04500 M AgNO3 for complete titration of both halides. If we assume there is x mol
of NaCl in the original sample, which of the following statements are CORRECT?
There are multiple correct answers.
MW(NaCl) = 58.44 g/mol. MW(KBr) = 119.002 g/mol.
There is x mol of KBr in the original sample.
x+(0.2386-58.44x)/119.002-50.00*0.04500
There is (0.002250 - x) mol of KBr in the original sample.
58.44x+119.002(0.002250-x)=0.2386

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