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Calculate the molarity of 198 g of barium iodide (Bal2) in 2.0 l of solution

Sagot :

Answer: The molarity of 198 g of barium iodide [tex](BaI_{2})[/tex] in 2.0 L of solution is 0.253 M.

Explanation:

Given: Mass = 198 g

Volume = 2.0 L

Molarity is the number of moles of solute present in liter of a solution.

Moles is the mass of a substance divided by its molar mass. So, moles of barium iodide (molar mass = 391.136 g/mol) is calculated as follows.

[tex]Moles = \frac{mass}{molar mass}\\= \frac{198 g}{391.136 g/mol}\\= 0.506 mol[/tex]

Now, molarity is calculated as follows.

[tex]Molarity = \frac{moles}{Volume (in L)}\\= \frac{0.506 mol}{2.0 L}\\= 0.253 M[/tex]

Thus, we can conclude that the molarity of 198 g of barium iodide [tex](BaI_{2})[/tex] in 2.0 L of solution is 0.253 M.

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