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What is the molarity of a solution prepared by dissolving 8 grams of BaCl2 in enough water to make a 450.0 mL solution?
A)0.17M

B)17.1M

C)0.085M

D)8.5M

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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