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What is the molarity of a solution formed by dissolving 97.7 g of LiBr in enough water to yield 1500.0 mL of solution?

Answer :

the molarity of the solution formed by dissolving 97.7 g of LiBr in enough water to yield 1500.0 mL of solution is approximately 0.750 M.

To calculate the molarity of a solution, we need to divide the moles of solute by the volume of the solution in liters.

First, let's calculate the moles of LiBr using the given mass and its molar mass:

Molar mass of LiBr:

Li: 6.941 g/mol

Br: 79.904 g/mol

Molar mass of LiBr = 6.941 g/mol + 79.904 g/mol = [tex]86.845 g/mol[/tex]

Moles of LiBr = [tex]Mass / Molar mass[/tex]

Moles of LiBr = 97.7 g / 86.845 g/mol

Next, we need to convert the volume of the solution from milliliters to liters:

[tex]Volume of the solution = 1500.0 mL = 1500.0 mL / 1000 mL/L = 1.500 L[/tex]

Now, we can calculate the molarity:

Molarity (M) = Moles of solute / Volume of solution (in liters)

Molarity = Moles of LiBr / Volume of solution

[tex]Molarity = (97.7 g / 86.845 g/mol) / 1.500 L[/tex]

Calculating this, we find:

Molarity ≈ [tex]0.750 M[/tex]

Therefore, the molarity of the solution formed by dissolving 97.7 g of LiBr in enough water to yield 1500.0 mL of solution is approximately 0.750 M.

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