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What mass of lithium phosphate (Li3PO4) is needed to prepare 500 mL (0.500 L) of a solution having a lithium ion concentration of 0.125 M?
a. 2.41 g
b. 6.75 g
c. 10.1 g
d. 19.3 g
e. 30.4 g

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Answer:

a. 2.41 g

Explanation:

First we calculate the number of lithium ion moles (Li⁺) in the solution, multiplying the concentration times the volume:

500 mL ⇒ 500 / 1000 = 0.500 L

  • 0.125 M * 0.500 L = 0.0625 mol Li⁺

Then we convert moles of Li⁺ to moles of Li₃PO₄, keeping in mind that for each mole of Li₃PO₄ there are 3 Li⁺ moles:

  • 0.0625 mol Li⁺ * [tex]\frac{1molLi_3PO_4}{3molLi^{+}}[/tex]= 0.02083 mol Li₃PO₄

Finally we convert Li₃PO₄ moles to grams, using its molar mass:

  • 0.02083 mol Li₃PO₄ * 115.79 g/mol = 2.41 g
Eduard22sly

The mass of lithium phosphate, Li₃PO₄ required to prepare the solution is 2.41 g

The correct answer to the question is Option A. 2.41 g

We'll begin by calculating the concentration of Li₃PO₄. This can be obtained as follow:

Li₃PO₄(aq) —> 3Li⁺(aq) + PO₄³¯(aq)

From the balanced equation above,

3 moles of Li⁺ is present in 1 mole of Li₃PO₄.

Therefore,

0.125 M Li⁺ will be present in = 0.125 / 3 = 0.0417 M Li₃PO₄.

Thus, the concentration of Li₃PO₄ is 0.0417 M

  • Next, we shall determine the number of mole of Li₃PO₄ in the solution.

Molarity of Li₃PO₄ = 0.0417 M

Volume = 0.5 L

Mole of Li₃PO₄ =?

Mole = Molarity x Volume

Mole of Li₃PO₄ = 0.0417 × 0.5

Mole of Li₃PO₄ = 0.02085 mole

  • Finally, we shall determine the mass of Li₃PO₄ required to prepare the solution.

Mole of Li₃PO₄ = 0.02085 mole

Molar mass of Li₃PO₄ = (3×7) + 31 + (16×4) = 116 g/mol

Mass of Li₃PO₄ =?

Mass = mole × molar mass

Mass of Li₃PO₄ = 0.02085 × 116

Mass of Li₃PO₄ = 2.41 g

Thus, the mass of Li₃PO₄ required to prepare the solution is 2.41 g

The correct answer to the question is Option A. 2.41 g

Learn more: https://brainly.com/question/3126258

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