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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]?

Answer :

Final answer:

The molarity of the H3PO4 solution is calculated by dividing the number of moles of H3PO4 by the volume of the solution in liters. After the required conversions are made, the molarity of the given H3PO4 solution is found to be 14.6M.

Explanation:

To calculate the molarity of H3PO4, we need to know the number of moles of H3PO4 and the volume of the solution in liters. Given that the solution is 85.5% H3PO4 by mass, this means that in 100 grams of the solution there are 85.5 grams of H3PO4. Using the molar mass of H3PO4 (~98.0 g/mol), we can convert grams to moles.

85.5 g H3PO4 * (1 mol/98.0 g) = 0.867 moles H3PO4

Given the density of the solution is 1.69 g/mL, or 1690 g/L, and we have 100 grams of solution, we can convert this to liters:

100g * (1L/1690g) = 0.0592 liters

Finally, to get molarity, we divide the number of moles by the volume in liters:

Molarity = Moles/Volume = 0.867 moles/0.0592L = 14.6 M

Therefore, the molarity of the H3PO4 solution is 14.6M.

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