Respuesta :
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
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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