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You need to make an aqueous solution of 0.238 M aluminum chloride for an experiment in lab, using a 250 mL volumetric flask. How much solid aluminum chloride should you add? grams

You need to make an aqueous solution of 0.238 M aluminum chloride for an experiment in lab, using a 250 mL volumetric flask. How much solid aluminum chloride should you add? grams

Chemistry
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
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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[References] Use the References to access important values if needed for this question. You need to make an aqueous solution of 0.238 M aluminum chloride for an experiment in lab, using a 250 mL volumetric flask. How much solid aluminum chloride should you add? grams
Transcribed Image Text:[References] Use the References to access important values if needed for this question. You need to make an aqueous solution of 0.238 M aluminum chloride for an experiment in lab, using a 250 mL volumetric flask. How much solid aluminum chloride should you add? grams
The bromide ion concentration in a solution may be determined by the precipitation of silver bromide. Ag+(aq) + Br (aq) —AgBr(s) A student finds that 21.18 mL of 0.7650 M silver nitrate is needed to precipitate all of the bromide ion in a 50.00-mL sample of an unknown. What is the molarity of the bromide ion in the student's unknown? M
Transcribed Image Text:The bromide ion concentration in a solution may be determined by the precipitation of silver bromide. Ag+(aq) + Br (aq) —AgBr(s) A student finds that 21.18 mL of 0.7650 M silver nitrate is needed to precipitate all of the bromide ion in a 50.00-mL sample of an unknown. What is the molarity of the bromide ion in the student's unknown? M
Expert Solution
Step 1

Given that,

Molarity of aluminium chloride=0.238M 

Normalityof aluminium chloride=molarity*n factor   (n=3)

N=0.238*3N=0.714N

Molecular weight of aluminium chloride=133.34 g/mol

Equivalent weight of aluminium chloride=133.34 g/mol/3=44.44g/mol 

Volume of the solution=250mL 

Amount of aluminium chloride is required(w)=?

 

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