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A concentrated phosphoric acid solution is 85.5% [tex]H_3PO_4[/tex] by mass and has a density of 1.69 g/mL at 25°C. What is the molarity of [tex]H_3PO_4[/tex]?

A. 14.7 M
B. 0.166 M
C. 5.16 M
D. 19.4 M
E. 0.0516 M

Answer :

Answer:

a. 14.7 M

Explanation:

Given that :

[tex]H_3PO_4[/tex] is 85.5 % w/w in solution.

It means that 85.5 g of the salt is present in 100 g of the solution.

Mass of solution = 100 g

Density = 1.69 g/mL

Volume of the solution = Mass / Density = 100 / 1.69 mL = 59.17 mL = 0.05917 L

Moles of [tex]H_3PO_4[/tex] can be calculated as:

Molar mass of [tex]H_3PO_4[/tex] = 97.994 g/mol

The formula for the calculation of moles is shown below:

[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]

Thus,

[tex]Moles= \frac{85.5\ g}{97.994\ g/mol}[/tex]

Moles = 0.8725 moles

Considering:

[tex]Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}[/tex]

Molarity = moles/ Volume of solution = 0.8725 / 0.05917 = 14.7 M

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Rewritten by : Barada

The molarity of the phosphoric acid, H₃PO₄ solution is 14.7 M

The correct answer to the question is Option A. 14.7 M

Let the mass of the solution be 100 g.

Therefore, the mass of 85.5% of H₃PO₄ is 85.5 g

  • Next, we shall determine the volume of the solution.

Density of solution = 1.69 g/mL

Mass of solution = 100 g

Volume of solution =?

Volume = mass / density

Volume of solution = 100 / 1.69

Volume of solution = 59.17 mL

Divide by 1000 to express in litre

Volume of solution = 59.17 / 1000

Volume of solution = 0.05917 L

  • Next, we shall determine the mole of H₃PO₄.

Mass of H₃PO₄ = 85.5 g

Molar mass of H₃PO₄ = (3×1) + 31 + (16×4) = 98 g/mol

Mole of H₃PO₄ =?

Mole = mass / molar mass

Mole of H₃PO₄ = 85.5 / 98

Mole of H₃PO₄ = 0.872 mole

  • Finally, we shall determine the molarity of the H₃PO₄ solution

Mole of H₃PO₄ = 0.872 mole

Volume of solution = 0.05917 L

Molarity of H₃PO₄ =?

Molarity = mole / Volume

Molarity of H₃PO₄ = 0.872 / 0.05917

Molarity of H₃PO₄ = 14.7 M

The correct answer to the question is Option A. 14.7 M

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