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Sagot :
The molarity of the solution prepared is 0.085 M
Stoichiometry
From the question, we are to determine the molarity of the solution
First, we will determine the number of moles of BaCl₂ present
Using the formula,
[tex]Number\ of\ moles = \frac{Mass}{Molar\ mass}[/tex]
Mass = 8 g
Molar mass of BaCl₂ = 208.23 g/mol
Then,
Number of moles of BaCl₂ present = [tex]\frac{8}{208.23}[/tex]
Number of moles of BaCl₂ present = 0.03842 moles
Now, for the molarity of the solution
[tex]Molarity = \frac{Number\ of\ moles}{Volume}[/tex]
From the given information,
Volume = 450.0 mL = 0.450 L
Molarity = [tex]\frac{0.03842}{0.450}[/tex]
Molarity = 0.085 M
Hence, the molarity of the solution prepared is 0.085 M
Learn more on Stoichiometry here: https://brainly.com/question/1566365
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