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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?
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Excess (NH4)2SO4 was added to a 65.0 mL solution containing BaCl2 (MW=...
**Solution:**
To find the molarity of BaCl2 in the solution, we need to first determine the number of moles of BaSO4 precipitate formed and then use stoichiometry to relate it to the number of moles of BaCl2.
**Step 1: Calculate the number of moles of BaSO4 precipitate formed**
The mass of the BaSO4 precipitate formed is given as 0.2791 g. We can use the formula weight of BaSO4 to calculate the number of moles.
Formula weight of BaSO4 = 233.43 g/mol
Number of moles of BaSO4 = mass / formula weight
= 0.2791 g / 233.43 g/mol
= 0.001195 mol
**Step 2: Use stoichiometry to relate the number of moles of BaSO4 to BaCl2**
From the balanced chemical equation of the reaction between BaCl2 and (NH4)2SO4, we can see that 1 mole of BaSO4 is formed for every 1 mole of BaCl2. This means that the number of moles of BaSO4 is equal to the number of moles of BaCl2.
Therefore, the molarity of BaCl2 in the solution can be calculated using the equation:
Molarity = moles / volume
We know the number of moles of BaSO4 from Step 1 (0.001195 mol) and the volume of the solution is given as 65.0 mL. However, the volume needs to be converted to liters in order to use the equation.
Volume = 65.0 mL = 65.0 mL * (1 L / 1000 mL) = 0.065 L
Molarity of BaCl2 = 0.001195 mol / 0.065 L
= 0.01838 M
Therefore, the molarity of BaCl2 in the solution is 0.01838 M.
To find the molarity of BaCl2 in the solution, we need to first determine the number of moles of BaSO4 precipitate formed and then use stoichiometry to relate it to the number of moles of BaCl2.
**Step 1: Calculate the number of moles of BaSO4 precipitate formed**
The mass of the BaSO4 precipitate formed is given as 0.2791 g. We can use the formula weight of BaSO4 to calculate the number of moles.
Formula weight of BaSO4 = 233.43 g/mol
Number of moles of BaSO4 = mass / formula weight
= 0.2791 g / 233.43 g/mol
= 0.001195 mol
**Step 2: Use stoichiometry to relate the number of moles of BaSO4 to BaCl2**
From the balanced chemical equation of the reaction between BaCl2 and (NH4)2SO4, we can see that 1 mole of BaSO4 is formed for every 1 mole of BaCl2. This means that the number of moles of BaSO4 is equal to the number of moles of BaCl2.
Therefore, the molarity of BaCl2 in the solution can be calculated using the equation:
Molarity = moles / volume
We know the number of moles of BaSO4 from Step 1 (0.001195 mol) and the volume of the solution is given as 65.0 mL. However, the volume needs to be converted to liters in order to use the equation.
Volume = 65.0 mL = 65.0 mL * (1 L / 1000 mL) = 0.065 L
Molarity of BaCl2 = 0.001195 mol / 0.065 L
= 0.01838 M
Therefore, the molarity of BaCl2 in the solution is 0.01838 M.
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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?
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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? 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? 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?.
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? 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? 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?.
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? 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?, 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? 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? theory, EduRev gives you an
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