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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.
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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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