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Chapter 13: Problem 42

A solution contains 32.7 \(\mathrm{g} \mathrm{H}_{3} \mathrm{PO}_{4}\) in 455 \(\mathrm{mL}\) of solution. Calculate its molarity.

Short Answer

Expert verified
The molarity of the solution is 0.734 M.

Step by step solution

01

Calculate the moles of phosphoric acid

To calculate the number of moles of \(\mathrm{H}_{3} \mathrm{PO}_{4}\), divide the given mass of the solution by its molar mass. The molar mass of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) is \(3 \times 1.01\, \mathrm{g/mol} + 15.99\, \mathrm{g/mol} + 4 \times 16.00\, \mathrm{g/mol} = 97.99\, \mathrm{g/mol}\). Therefore, the number of moles becomes \(32.7\, \mathrm{g} / 97.99\, g/mol = 0.334\, mol\).
02

Convert the volume from milliliters to liters

The volume of the solution in liters is calculated by dividing the given volume in milliliters by 1000. Hence, the volume of the solution is \(455\, ml / 1000 = 0.455\, L\).
03

Calculate the molarity

The molarity of the solution can now be calculated using the formula: Molarity = moles of solute / liters of solution. Substituting the earlier calculated values becomes: \(0.334\, mol / 0.455\, L = 0.734\, M\).

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