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Chapter 13: Problem 20

A typical commercial grade aqueous phosphoric acid is \(75 \% \mathrm{H}_{3} \mathrm{PO}_{4}\) by mass and has a density of \(1.57 \mathrm{g} / \mathrm{mL}\) What is the molarity of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) in this solution?

Short Answer

Expert verified
The molarity of H3PO4 in the solution is 11.95 M.

Step by step solution

01

Calculation of Mass of the Solute

Extract 100g of the solution since it's more practical to calculate with base 100% when dealing with percentages. Given this, the solution consists of 75g H3PO4 (from the 75% w/w information)
02

Calculate the moles of H3PO4

To convert grams to moles, the molar mass of H3PO4 is required. It can be calculated as \(3 \times 1.01g/mol (H) + 15.99 g/mol (P) + 4 \times 16.00g/mol (O) = 97.99g/mol\). Thus, the moles of H3PO4 in 75g is \( \frac{75 g}{97.99 g/mol} = 0.765 mol\)
03

Calculate the volume of the solution in liters

With the given density of 1.57 g/mL, convert this density to g/L to obtain \(1.57 g/mL \times 1000 = 1570 g/L \). Then, find the volume occupied by 100g of the solution by using the formula 'mass = volume \times density' rearranged to 'volume = mass / density' giving \( \frac{100g}{1570 g/L} = 0.064 L \)
04

Compute the Molarity

Finally, molarity can be calculated by substituting the obtained values into the molarity formula: \( Molarity = \frac{Amount of Solute (moles)}{Volume of solution (L)} = \frac{0.765mol}{0.064L} \)= 11.95 M.

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