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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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