Respuesta :

superman1987

Answer:

1.5 M.

Explanation:

  • Molarity (M) is defined as the no. of moles of solute dissolved in a 1.0 L of the solution.

M = (no. of moles of LiBr)/(Volume of the solution (L).

∵ no. of moles of LiBr = (mass/molar mass) of LiBr = (97.7 g)/(86.845 g/mol) = 1.125 mol.

Volume of the solution = 750.0 mL = 0.75 L.

∴ M = (no. of moles of luminol)/(Volume of the solution (L) = (1.125 mol)/(0.75 L) = 1.5 M.

dexteright02

Hello!

Determine the molarity of a solution formed by dissolving 97.7 g LiBr in enough water to yield 750.0 ml of solution.

We have the following data:  

M (Molarity) =? (in mol / L)

m1 (mass of the solute) = 97.7 g

V (solution volume) = 750 ml → V (solution volume) = 0.75 L

MM (molar mass of LiBr)

Li = 6.941 u

Br = 79.904 u

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MM (molar mass of LiBr) = 6.941   + 79.904

MM (molar mass of LiBr) = 86.845 g/mol

Now, let's apply the data to the formula of Molarity, let's see:

[tex]M = \dfrac{m_1}{MM*V}[/tex]

[tex]M = \dfrac{97.7}{86.845*0.75}[/tex]

[tex]M = \dfrac{97.7}{65.13375}[/tex]

[tex]M = 1.49999... \to \boxed{\boxed{M \approx 1.5\:mol/L}}\:\:\:\:\:\:\bf\green{\checkmark}[/tex]

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*** Another way to solve is to find the number of moles (n1) and soon after finding the molarity (M), let's see:

[tex]n_1 = \dfrac{m_1\:(g)}{MM\:(g/mol)}[/tex]

[tex]n_1 = \dfrac{97.7\:\diagup\!\!\!\!\!g}{86.845\:\diagup\!\!\!\!\!g/mol}[/tex]

[tex]n = 1.12499.. \to \boxed{n_1 \approx 1.125\:mol}[/tex]

[tex]M = \dfrac{n_1\:(mol)}{V\:(L)}[/tex]

[tex]M = \dfrac{1.125\:mol}{0.75\:L}[/tex]

[tex]\boxed{\boxed{M = 1.5\:mol/L}}\:\:\:\:\:\:\bf\blue{\checkmark}[/tex]

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[tex]\bf\purple{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}\:\:\ddot{\smile}[/tex]