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Chapter 11: Problem 29

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
The density of the phosphoric acid solution is 1.058 g/mL, the mole fraction of H₃PO₄ is 0.0180, the molarity of the solution is 0.981 M, and the molality is 1.02 mol/kg.

Step by step solution

01

1. Calculate the moles of H₃PO₄

First, we need to calculate the number of moles of H₃PO₄ in the solution. We will use the formula: moles = mass of solute / molar mass. The molar mass of H₃PO₄ = 3(1.01) + 1(15.99) + 4(16.00) = 97.99 g/mol. So, moles of H₃PO₄ = \( \frac{10.0 \ g}{97.99 \ g/mol} = 0.102 \ \text{moles} \)
02

2. Calculate the mass of water and the mass of the solution

We are given that the mass of H₃PO₄ was 10.0 g and the volume of water is 100.0 mL. Assuming the density of water is 1.00 g/cm³, we can convert the volume of water to mass: Mass of water = density × volume = 1.00 g/cm³ × 100.0 cm³ = 100.0 g The total mass of the solution will be the sum of the mass of H₃PO₄ and the mass of water: Total mass of solution = 10.0 g + 100.0 g = 110.0 g
03

3. Calculate the density of the solution

We are given that the volume of the solution is 104 mL. To find the density, we will use the formula: density = mass / volume. So, density of the solution = \( \frac{110.0 \ g}{104.0 \ mL} = 1.058 \ \frac{g}{mL} \)
04

4. Calculate the mole fraction of H₃PO₄

To find the mole fraction of H₃PO₄, we need to find the moles of water and then use the formula: mole_fraction = moles of solute / (moles of solute + moles of solvent). First, we find the moles of water using the formula: moles = mass of solvent / molar mass. The molar mass of water = 18.015 g/mol. Moles of water = \( \frac{100.0 \ g}{18.015 \ g/mol} = 5.553 \ \text{moles} \) Now, we find the mole fraction of H₃PO₄: Mole fraction of H₃PO₄ = \( \frac{0.102 \ \text{moles}}{0.102 \ \text{moles} + 5.553 \ \text{moles}} = 0.0180 \)
05

5. Calculate the molarity of the solution

Molarity is defined as the number of moles of solute per liter of solution. We know that the volume of the solution is 104 mL. We can convert this into liters: Volume of solution = 104 mL × \( \frac{1 \ \text{L}}{1000 \ \text{mL}} \) = 0.104 L Now, we find the molarity using the formula: molarity = moles of solute / volume of solution (in liters): Molarity of H₃PO₄ = \( \frac{0.102 \ \text{moles}}{0.104 \ L} \) = 0.981 M
06

6. Calculate the molality of the solution

Molality is defined as the number of moles of solute per kilogram of solvent. We have already found the moles of H₃PO₄ (0.102 moles) and we know the mass of water (100.0 g). We can convert the mass of water to kilograms: Mass of water = 100.0 g × \( \frac{1 \ \text{kg}}{1000 \ \text{g}} \) = 0.100 kg Now, we find the molality using the formula: molality = moles of solute / mass of solvent (in kg): Molality of H₃PO₄ = \( \frac{0.102 \ \text{moles}}{0.100 \ \text{kg}} \) = 1.02 mol/kg In conclusion, the density of the solution is 1.058 g/mL, the mole fraction of H₃PO₄ is 0.0180, the molarity is 0.981 M, and the molality is 1.02 mol/kg.

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