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Chapter 11: Problem 37

A solution of phosphoric acid was made by dissolving 10.0 \(\mathrm{g}\) \(\mathrm{H}_{3} \mathrm{PO}_{4}\) in 100.0 \(\mathrm{mL}\) water. The resulting volume was 104 \(\mathrm{mL}\) . Calculate the density, mole fraction, molarity, and molality of the solution. Assume water has a density of 1.00 $\mathrm{g} / \mathrm{cm}^{3}$ .

Short Answer

Expert verified
In summary, the phosphoric acid solution has a density of 1.058 g/mL, a mole fraction of H3PO4 at 0.01804 and H2O at 0.9819, a molarity of 0.981 mol/L, and a molality of 1.02 mol/kg.

Step by step solution

01

Calculate moles of H3PO4 and H2O

First, we find the moles of H3PO4 (phosphoric acid) in the solution. The molecular weight of H3PO4 is: 3(1.008) + 1(30.97) + 4(16.00) = 97.988 g/mol Given that we have 10.0 g of H3PO4, the number of moles can be calculated as follows: Moles of H3PO4 = \(\frac{10.0 \, g}{97.988 \, g/mol}\) = 0.102 moles (approximately) Next, we calculate the moles of H2O. The molecular weight of H2O is: 2(1.008) + 1(16.00) = 18.016 g/mol Since the solution has a volume of 100 mL, the mass of H2O can be calculated using its density: Mass of H2O = volume × density = 100.0 mL × 1.00 g/mL = 100.0 g Moles of H2O = \(\frac{100.0 \, g}{18.016 \, g/mol}\) = 5.555 moles (approximately)
02

Find the mass of the solution

To determine the mass of the solution, we add the masses of H3PO4 and H2O: Mass of the solution = mass of H3PO4 + mass of H2O = 10.0 g + 100.0 g = 110.0 g
03

Calculate the density of the solution

The density of the solution can be calculated using the formula: Density = \(\frac{mass}{volume}\) In this case, we have: Density = \(\frac{110.0 \, g}{104 \, mL}\) = 1.058 g/mL (approximately)
04

Determine the mole fraction of H3PO4 and H2O

Mole fraction is calculated as the ratio of moles of a component to the total moles in the solution: Mole fraction of H3PO4 = \(\frac{moles \, of \, H_3PO_4}{total \, moles \, in \, solution}\) = \(\frac{0.102 \, mol}{0.102 \, mol + 5.555 \, mol}\) = 0.01804 (approximately) Mole fraction of H2O = \(\frac{moles \, of \, H_2O}{total \, moles \, in \, solution}\) = \(\frac{5.555 \, mol}{0.102 \, mol + 5.555 \, mol}\) = 0.9819 (approximately)
05

Find the molarity of the solution

Molarity is determined by dividing the moles of solute (H3PO4) by the volume of the solution (in liters): Molarity = \(\frac{moles \, of \, H_3PO_4}{volume \, of \, solution}\) = \(\frac{0.102 \, mol}{0.104 \, L}\) = 0.981 mol/L (approximately)
06

Calculate the molality of the solution

Molality is defined as the moles of solute per kilogram of solvent (H2O): Molality = \(\frac{moles \, of \, H_3PO_4}{mass \, of \, H_2O}\) = \(\frac{0.102 \, mol}{0.1000 \, kg}\) = 1.02 mol/kg (approximately) In summary, the phosphoric acid solution has a density of 1.058 g/mL, a mole fraction of H3PO4 at 0.01804 and H2O at 0.9819, a molarity of 0.981 mol/L, and a molality of 1.02 mol/kg.

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