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Chapter 13: Problem 47

A sulfuric acid solution containing \(697.6 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) per liter of solution has a density of \(1.395 \mathrm{~g} / \mathrm{cm}^{3} .\) Calculate (a) the mass percentage, (b) the mole fraction, (c) the molality, \((\mathbf{d})\) the molarity of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) in this solution.

Short Answer

Expert verified
(a) 50.03%, (b) 0.155, (c) 10.21 mol/kg, (d) 7.12 M

Step by step solution

01

Calculate Total Mass of Solution per Liter

The density is given as \(1.395 \ \text{g/cm}^3\), which is equivalent to \(1.395 \ \text{g/ml}\). Since there are 1000 ml in a liter, the mass of 1 liter of solution is \(1.395 \times 1000 = 1395 \ \text{g}\).
02

Calculate Mass Percentage of H2SO4

The mass of \(\text{H}_2\text{SO}_4\) in the solution is given as \(697.6 \ \text{g/l}\). The mass percentage is calculated as: \[\text{Mass Percentage} = \left( \frac{697.6}{1395} \right) \times 100 = 50.03\%\]
03

Calculate Molar Mass of H2SO4

The molar mass of \(\text{H}_2\text{SO}_4\) is calculated by adding the atomic masses of its constituent atoms: 2 hydrogens, 1 sulfur, and 4 oxygens: \[ \text{Molar Mass} = (2\times1) + 32 + (4\times16) = 2 + 32 + 64 = 98 \ \text{g/mol}\]
04

Calculate Moles of H2SO4 in the Solution

Using the molar mass from the previous step, calculate the moles of \(\text{H}_2\text{SO}_4\) in the solution. \[ \text{Moles of } \text{H}_2\text{SO}_4 = \frac{697.6}{98} = 7.12 \ \text{mol}\]
05

Calculate Mole Fraction of H2SO4

First, calculate the moles of water in the solution by finding the mass of water and dividing by its molar mass. The mass of water is the total mass of the solution minus the mass of \(\text{H}_2\text{SO}_4\).\[ \text{Mass of Water} = 1395 - 697.6 = 697.4 \ \text{g}\]\[ \text{Moles of Water} = \frac{697.4}{18} = 38.74 \ \text{mol}\]Calculate the mole fraction of \(\text{H}_2\text{SO}_4\):\[ \text{Mole Fraction of H}_2\text{SO}_4 = \frac{7.12}{7.12 + 38.74} = 0.155\]
06

Calculate Molality of H2SO4

Molality is the moles of solute per kilogram of solvent. Calculate the molality of \(\text{H}_2\text{SO}_4\) using the moles of \(\text{H}_2\text{SO}_4\) and the mass of water in kg:\[ \text{Molality} = \frac{7.12}{0.6974} = 10.21 \ \text{mol/kg}\]
07

Calculate Molarity of H2SO4

Given that the solution is already 1 liter, the molarity is simply the moles of \(\text{H}_2\text{SO}_4\) per liter of solution. Thus, the molarity is:\[ \text{Molarity} = 7.12 \ \text{M}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molarity
Molarity is a way to express the concentration of a solute in a solution. It tells us how many moles of solute are present in one liter of solution.
To calculate molarity (M), you use the formula:
  • \[ \text{Molarity} = \frac{\text{moles of solute}}{\text{liters of solution}} \]
For example, in the case of sulfuric acid, if we have 7.12 moles of H\(_2\)SO\(_4\) in one liter, the molarity is 7.12 M. Knowing molarity is crucial when preparing solutions for reactions, as it allows precise control over the quantities involved.
Molality
Molality is another way to express the concentration of a solution, but unlike molarity, it depends on the mass of the solvent rather than the volume of the solution.
The formula for molality (m) is as follows:
  • \[ \text{Molality} = \frac{\text{moles of solute}}{\text{kilograms of solvent}} \]
In our sulfuric acid example, the molarity was 10.21 mol/kg, meaning that there are 10.21 moles of H\(_2\)SO\(_4\) in every kilogram of water. Molality is particularly useful when dealing with temperature and pressure changes because it solely depends on mass, not volume, which can vary with conditions.
Mole Fraction
Mole fraction is a measure of the concentration of a component in a mixture.
It is calculated using the ratio:
  • \[ \text{Mole Fraction} = \frac{\text{moles of component}}{\text{total moles in solution}} \]
For sulfuric acid in a solution, the mole fraction of H\(_2\)SO\(_4\) is calculated by dividing the moles of H\(_2\)SO\(_4\) by the total moles of all substances in the solution. In this case, the mole fraction is 0.155, indicating that H\(_2\)SO\(_4\) makes up 15.5% of the solution in terms of mole count. This measurement is particularly helpful in thermodynamic calculations.
Mass Percentage
Mass percentage represents how much of a particular substance is present in a mixture compared to the whole mixture.
It's given by:
  • \[ \text{Mass Percentage} = \left( \frac{\text{mass of solute}}{\text{total mass of solution}} \right) \times 100 \]
In the case of our sulfuric acid solution, the mass percentage is 50.03%. This means that 50.03% of the mass of the entire solution is made up of H\(_2\)SO\(_4\). Mass percentage is widely used in chemistry for expressing the concentration of mixtures and can be particularly useful in situations where precise solution masses are important.

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Most popular questions from this chapter

The density of acetonitrile \(\left(\mathrm{CH}_{3} \mathrm{CN}\right)\) is \(0.786 \mathrm{~g} / \mathrm{mL}\) and the density of methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) is \(0.791 \mathrm{~g} / \mathrm{mL}\). A solution is made by dissolving \(25.0 \mathrm{~mL}\) of \(\mathrm{CH}_{3} \mathrm{OH}\) in \(100 \mathrm{~mL}\) of \(\mathrm{CH}_{3} \mathrm{CN}\). (a) What is the mole fraction of methanol in the solution? (b) What is the molality of the solution? (c) Assuming that the volumes are additive, what is the molarity of \(\mathrm{CH}_{3} \mathrm{OH}\) in the solution?

(a) Do colloids made only of gases exist? Why or why not? (b) In the 1850 s, Michael Faraday prepared ruby-red colloids of gold nanoparticles in water that are still stable today. These brightly colored colloids look like solutions. What experiment(s) could you do to determine whether a given colored preparation is a solution or colloid?

A lithium salt used in lubricating grease has the formula \(\mathrm{LiC}_{n} \mathrm{H}_{2 n+1} \mathrm{O}_{2}\). The salt is soluble in water to the extent of \(0.036 \mathrm{~g}\) per \(100 \mathrm{~g}\) of water at \(25^{\circ} \mathrm{C}\). The osmotic pressure of this solution is found to be \(7.61 \mathrm{kPa}\). Assuming that molality and molarity in such a dilute solution are the same and that the lithium salt is completely dissociated in the solution, determine an appropriate value of \(n\) in the formula for the salt.

Choose the best answer: A colloidal dispersion of one liquid in another is called \((\mathbf{a})\) a gel, \((\mathbf{b})\) an emulsion, (c) a foam, (d) an aerosol.

The Baltic Sea has a salinity of \(1.0 \%\), that is, its water contains \(10 \mathrm{~g}\) of salt for every liter of solution. Assuming that the solute consists entirely of \(\mathrm{NaCl}\) (in fact, over \(90 \%\) of the salt is indeed \(\mathrm{NaCl}\), calculate the osmotic pressure of this seawater at \(15^{\circ} \mathrm{C}\).
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