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Excess (NH4)2SO4 was added to a 65.0 mL solution containing BaCl2 (MW=208.23). The resulting BaSO4 (MW=233.43 g/mol) precipitate had a mass of 0.2791 g after it was filtered and dried. What is the molarity of BaCl2 in the solution? I found my answer to be 0.00183945 and rounded to 0.00184 and it is incorrect. Please help?
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Excess (NH4)2SO4 was added to a 65.0 mL solution containing BaCl2 (MW=...
**Given:**
- Mass of BaSO4 precipitate = 0.2791 g
- Molecular weight of BaSO4 = 233.43 g/mol
- Volume of solution = 65.0 mL
- Molecular weight of BaCl2 = 208.23 g/mol
**To find:**
Molarity of BaCl2 in the solution
**Solution:**
Step 1: Calculate the number of moles of BaSO4 precipitate
- Moles of BaSO4 = mass / molecular weight
- Moles of BaSO4 = 0.2791 g / 233.43 g/mol
Step 2: Calculate the number of moles of BaCl2
- Since BaSO4 and BaCl2 are in a 1:1 ratio, the number of moles of BaCl2 is the same as the number of moles of BaSO4.
Step 3: Calculate the volume of the solution in liters
- Volume of the solution = 65.0 mL = 65.0 mL * (1 L / 1000 mL)
Step 4: Calculate the molarity of BaCl2
- Molarity = moles / volume
- Molarity = moles of BaCl2 / volume of solution
**Summary:**
The molarity of BaCl2 in the solution can be calculated by dividing the number of moles of BaCl2 by the volume of the solution in liters.
- Mass of BaSO4 precipitate = 0.2791 g
- Molecular weight of BaSO4 = 233.43 g/mol
- Volume of solution = 65.0 mL
- Molecular weight of BaCl2 = 208.23 g/mol
**To find:**
Molarity of BaCl2 in the solution
**Solution:**
Step 1: Calculate the number of moles of BaSO4 precipitate
- Moles of BaSO4 = mass / molecular weight
- Moles of BaSO4 = 0.2791 g / 233.43 g/mol
Step 2: Calculate the number of moles of BaCl2
- Since BaSO4 and BaCl2 are in a 1:1 ratio, the number of moles of BaCl2 is the same as the number of moles of BaSO4.
Step 3: Calculate the volume of the solution in liters
- Volume of the solution = 65.0 mL = 65.0 mL * (1 L / 1000 mL)
Step 4: Calculate the molarity of BaCl2
- Molarity = moles / volume
- Molarity = moles of BaCl2 / volume of solution
**Summary:**
The molarity of BaCl2 in the solution can be calculated by dividing the number of moles of BaCl2 by the volume of the solution in liters.
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Excess (NH4)2SO4 was added to a 65.0 mL solution containing BaCl2 (MW=208.23). The resulting BaSO4 (MW=233.43 g/mol) precipitate had a mass of 0.2791 g after it was filtered and dried. What is the molarity of BaCl2 in the solution? I found my answer to be 0.00183945 and rounded to 0.00184 and it is incorrect. Please help?
Question Description
Excess (NH4)2SO4 was added to a 65.0 mL solution containing BaCl2 (MW=208.23). The resulting BaSO4 (MW=233.43 g/mol) precipitate had a mass of 0.2791 g after it was filtered and dried. What is the molarity of BaCl2 in the solution? I found my answer to be 0.00183945 and rounded to 0.00184 and it is incorrect. Please help? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about Excess (NH4)2SO4 was added to a 65.0 mL solution containing BaCl2 (MW=208.23). The resulting BaSO4 (MW=233.43 g/mol) precipitate had a mass of 0.2791 g after it was filtered and dried. What is the molarity of BaCl2 in the solution? I found my answer to be 0.00183945 and rounded to 0.00184 and it is incorrect. Please help? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Excess (NH4)2SO4 was added to a 65.0 mL solution containing BaCl2 (MW=208.23). The resulting BaSO4 (MW=233.43 g/mol) precipitate had a mass of 0.2791 g after it was filtered and dried. What is the molarity of BaCl2 in the solution? I found my answer to be 0.00183945 and rounded to 0.00184 and it is incorrect. Please help?.
Excess (NH4)2SO4 was added to a 65.0 mL solution containing BaCl2 (MW=208.23). The resulting BaSO4 (MW=233.43 g/mol) precipitate had a mass of 0.2791 g after it was filtered and dried. What is the molarity of BaCl2 in the solution? I found my answer to be 0.00183945 and rounded to 0.00184 and it is incorrect. Please help? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about Excess (NH4)2SO4 was added to a 65.0 mL solution containing BaCl2 (MW=208.23). The resulting BaSO4 (MW=233.43 g/mol) precipitate had a mass of 0.2791 g after it was filtered and dried. What is the molarity of BaCl2 in the solution? I found my answer to be 0.00183945 and rounded to 0.00184 and it is incorrect. Please help? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Excess (NH4)2SO4 was added to a 65.0 mL solution containing BaCl2 (MW=208.23). The resulting BaSO4 (MW=233.43 g/mol) precipitate had a mass of 0.2791 g after it was filtered and dried. What is the molarity of BaCl2 in the solution? I found my answer to be 0.00183945 and rounded to 0.00184 and it is incorrect. Please help?.
Solutions for Excess (NH4)2SO4 was added to a 65.0 mL solution containing BaCl2 (MW=208.23). The resulting BaSO4 (MW=233.43 g/mol) precipitate had a mass of 0.2791 g after it was filtered and dried. What is the molarity of BaCl2 in the solution? I found my answer to be 0.00183945 and rounded to 0.00184 and it is incorrect. Please help? in English & in Hindi are available as part of our courses for Chemistry.
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Excess (NH4)2SO4 was added to a 65.0 mL solution containing BaCl2 (MW=208.23). The resulting BaSO4 (MW=233.43 g/mol) precipitate had a mass of 0.2791 g after it was filtered and dried. What is the molarity of BaCl2 in the solution? I found my answer to be 0.00183945 and rounded to 0.00184 and it is incorrect. Please help?, a detailed solution for Excess (NH4)2SO4 was added to a 65.0 mL solution containing BaCl2 (MW=208.23). The resulting BaSO4 (MW=233.43 g/mol) precipitate had a mass of 0.2791 g after it was filtered and dried. What is the molarity of BaCl2 in the solution? I found my answer to be 0.00183945 and rounded to 0.00184 and it is incorrect. Please help? has been provided alongside types of Excess (NH4)2SO4 was added to a 65.0 mL solution containing BaCl2 (MW=208.23). The resulting BaSO4 (MW=233.43 g/mol) precipitate had a mass of 0.2791 g after it was filtered and dried. What is the molarity of BaCl2 in the solution? I found my answer to be 0.00183945 and rounded to 0.00184 and it is incorrect. Please help? theory, EduRev gives you an
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