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You need to make an aqueous solution of 0.177 M magnesium bromide for an experiment in lab, using a 300 mL volumetric flask. How much solid magnesium bromide should you add? 15.66 grams

You need to make an aqueous solution of 0.177 M magnesium bromide for an experiment in lab, using a 300 mL volumetric flask. How much solid magnesium bromide should you add? 15.66 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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**Preparation of Aqueous Solution for Laboratory Experiment** **Problem Statement:** You need to make an aqueous solution of 0.177 M magnesium bromide for an experiment in the lab, using a 300 mL volumetric flask. How much solid magnesium bromide should you add? **Answer:** You should add 15.66 grams of solid magnesium bromide. This calculation involves determining the amount of magnesium bromide (MgBr₂) in grams needed to achieve a specified molarity (M) in a given volume of solution. The data provided indicates that the desired concentration is 0.177 M in a 300 mL solution. The resulting mass of magnesium bromide required is 15.66 grams. This information is crucial for accurately preparing the solution for laboratory experiments to ensure consistency and reliable results.
Transcribed Image Text:**Preparation of Aqueous Solution for Laboratory Experiment** **Problem Statement:** You need to make an aqueous solution of 0.177 M magnesium bromide for an experiment in the lab, using a 300 mL volumetric flask. How much solid magnesium bromide should you add? **Answer:** You should add 15.66 grams of solid magnesium bromide. This calculation involves determining the amount of magnesium bromide (MgBr₂) in grams needed to achieve a specified molarity (M) in a given volume of solution. The data provided indicates that the desired concentration is 0.177 M in a 300 mL solution. The resulting mass of magnesium bromide required is 15.66 grams. This information is crucial for accurately preparing the solution for laboratory experiments to ensure consistency and reliable results.
--- ### Dissociation Reaction of Chromium(III) Acetate in Water The compound **chromium(III) acetate** is a strong electrolyte. The task requires writing the reaction when solid chromium(III) acetate is put into water. **Problem Statement:** Write the reaction when solid **chromium(III) acetate** is put into water. - Include states of matter in your answer. **Given Reactant:** \[ \text{Cr(CH}_3\text{COO)}_3 \] **Dissociation Reaction:** \[ \text{Cr(CH}_3\text{COO)}_3 (s) \rightarrow \text{Cr}^{3+} (aq) + 3\text{CH}_3\text{COO}^- (aq) \] **Explanation:** - **Cr(CH₃COO)₃ (s)**: Indicating solid chromium(III) acetate. - **Cr³⁺ (aq)**: Chromium ions in aqueous form, after the dissociation. - **3CH₃COO⁻ (aq)**: Three acetate ions in aqueous form. The solid chromium(III) acetate dissociates completely into chromium ions and acetate ions in water, which is characteristic of a strong electrolyte. ---
Transcribed Image Text:--- ### Dissociation Reaction of Chromium(III) Acetate in Water The compound **chromium(III) acetate** is a strong electrolyte. The task requires writing the reaction when solid chromium(III) acetate is put into water. **Problem Statement:** Write the reaction when solid **chromium(III) acetate** is put into water. - Include states of matter in your answer. **Given Reactant:** \[ \text{Cr(CH}_3\text{COO)}_3 \] **Dissociation Reaction:** \[ \text{Cr(CH}_3\text{COO)}_3 (s) \rightarrow \text{Cr}^{3+} (aq) + 3\text{CH}_3\text{COO}^- (aq) \] **Explanation:** - **Cr(CH₃COO)₃ (s)**: Indicating solid chromium(III) acetate. - **Cr³⁺ (aq)**: Chromium ions in aqueous form, after the dissociation. - **3CH₃COO⁻ (aq)**: Three acetate ions in aqueous form. The solid chromium(III) acetate dissociates completely into chromium ions and acetate ions in water, which is characteristic of a strong electrolyte. ---
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