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Answer: There are 0.5 grams of barium sulfate are present in 250 of 2.0 M [tex]BaSO_{4}[/tex] solution.
Explanation:
Given: Molarity of solution = 2.0 M
Volume of solution = 250 mL
Convert mL int L as follows.
[tex]1 mL = 0.001 L\\250 mL = 250 mL \times \frac{0.001 L}{1 mL}\\= 0.25 L[/tex]
Molarity is the number of moles of solute present in liter of solution. Hence, molarity of the given [tex]BaSO_{4}[/tex] solution is as follows.
[tex]Molarity = \frac{mass}{Volume (in L)}\\2.0 M = \frac{mass}{0.25 L}\\mass = 0.5 g[/tex]
Thus, we can conclude that there are 0.5 grams of barium sulfate are present in 250 of 2.0 M [tex]BaSO_{4}[/tex] solution.
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