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if 20.1 g of KBr (MM=119.00 g/mol) are added to a 500.0 mL volumetric flask and water is added to fill the flask what is the concentration of KBr in the resulting solution

Sagot :

The concentration of KBr in the resultion solution is 0.34 M

Stoichiometry

From the question, we are to determine the concentration of KBr in the resulting solution

First, we will determine the number of mole of KBr present

Using the formula,

[tex]Number\ of\ moles = \frac{Mass}{molar\ mass}[/tex]

Mass of KBr = 20.1 g

Molar mass of KBr = 119.00 g/mol

∴ Number of moles of KBr present = [tex]\frac{20.1}{119.00}[/tex]
Number of moles of KBr present = 0.1689 mole

Now, for the concentration of the KBr

Using the formula,

[tex]Concentration = \frac{Number\ of\ moles}{Volume}[/tex]

Volume of solution = 500.0 mL = 0.50 L

∴ Concentration of KBr = [tex]\frac{0.1689}{0.50}[/tex]

Concentration of KBr = 0.3378 M

Concentration of KBr ≅ 0.34 M

Hence, the concentration of KBr in the resultion solution is 0.34 M

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