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Chapter 7: Problem 19

Calculate the mole fraction of each component in the following solutions: (a) \(2.5 .0 \mathrm{~g}\) of water and \(50.0 \mathrm{~g}\) of ethanol, \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{OH}\); (b) \(25.0 \mathrm{~g}\) of water and \(50.0 \mathrm{~g}\) of methanol, \(\mathrm{CH}_{3} \mathrm{OH}\); (c) a glucowe solution that is \(0.10 \mathrm{~m}\) \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)\).

Short Answer

Expert verified
Ethanol-water solution: mole fraction water = 2.5 / 18.02 / (2.5 / 18.02 + 50 / 46.07), mole fraction ethanol = 50 / 46.07 / (2.5 / 18.02 + 50 / 46.07). Methanol-water solution: mole fraction water = 25 / 18.02 / (25 / 18.02 + 50 / 32.04), mole fraction methanol = 50 / 32.04 / (25 / 18.02 + 50 / 32.04). Glucose solution (approximation for very dilute solutions): mole fraction glucose 鈮� 0.10, mole fraction water 鈮� 1 - 0.10.

Step by step solution

01

Calculate Molar Masses

Find the molar mass of each component by summing the molar masses of its constituent atoms. Water has a molar mass of 18.02 g/mol. For ethanol, C2H5OH, the molar mass is 2(12.01 g/mol) + 6(1.008 g/mol) + 16.00 g/mol + 1.008 g/mol, totaling 46.07 g/mol. For methanol, CH3OH, the molar mass is 12.01 g/mol + 4(1.008 g/mol) + 16.00 g/mol + 1.008 g/mol, totaling 32.04 g/mol. For glucose, C6H12O6, the molar mass is 6(12.01 g/mol) + 12(1.008 g/mol) + 6(16.00 g/mol), totaling 180.16 g/mol.
02

Convert Mass to Moles

Use the molar mass to convert grams to moles for each component. For water and ethanol: moles water = 2.5 g / 18.02 g/mol, moles ethanol = 50.0 g / 46.07 g/mol. For water and methanol: moles water = 25.0 g / 18.02 g/mol, moles methanol = 50.0 g / 32.04 g/mol. For the glucose solution, the concentration 0.10 m (molality) already gives the moles of glucose in 1 kg of water, there are 0.10 moles glucose per kilogram of water, and we assume the mass of water to be enough to consider its mole fraction nearly 1.
03

Calculate the Mole Fractions

The mole fraction of a component is defined as its number of moles divided by the total moles of all components. For component A: mole fraction A = moles A / (moles A + moles B). For each solution, calculate accordingly. For example, in mixed water and ethanol mixture: mole fraction water = moles water / (moles water + moles ethanol). Repeat for each component in every solution.
04

Perform the Calculations

Perform the calculations using the formula from Step 3. If these were the values calculated in Step 2: moles water = x, moles ethanol = y, then mole fraction water = x / (x + y) and mole fraction ethanol = y / (x + y). Use the actual values obtained from Step 2 for the calculations.
05

Report the Mole Fractions

After performing the above calculations for each component in all solutions, give the final mole fractions as the solution to the problem.

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Key Concepts

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

Molar Mass
Understanding the concept of molar mass is vital for a range of chemistry calculations, including the determination of mole fractions. Molar mass is defined as the mass (in grams) of one mole of any substance. A mole is a unit of measurement used in chemistry to express amounts of a chemical substance, represented by Avogadro's number, which is approximately 6.022 x 1023 particles (atoms, molecules, ions or electrons).

To calculate molar mass, we sum the atomic masses of all the atoms in a molecule. These atomic masses are found on the periodic table, usually beneath the symbol for each element, and are measured in atomic mass units (u). For example, the molar mass of water (H2O) is obtained by adding the atomic masses of two hydrogen atoms (1.008 g/mol each) and one oxygen atom (16.00 g/mol), giving us 18.02 g/mol.

Example:

In the case of ethanol (C2H5OH), we calculate molar mass by adding the molar masses of two carbon atoms, six hydrogen atoms, and one oxygen atom. This addition gives us the molar mass of ethanol as 46.07 g/mol.
Conversion of Mass to Moles
Conversion of mass to moles is a fundamental step in chemical calculations and is used to compare quantities of different substances on a common scale. To convert mass to moles, we simply divide the mass of the substance by its molar mass.

The formula for this conversion is:
\[\begin{equation} \text{Number of moles} = \frac{\text{Mass of substance (g)}}{\text{Molar mass (g/mol)}} \end{equation}\]
Looking at our earlier example, if we have 50.0 g of ethanol, to find the number of moles we divide by its molar mass, yielding:\[\begin{equation} \text{Number of moles ethanol} = \frac{50.0 \text{ g}}{46.07 \text{ g/mol}} \end{equation}\]
Carrying out this calculation allows us to express the mass of ethanol as a quantity of moles, which can then be used to find the mole fraction in a mixture.
Solution Concentration
Solution concentration often refers to the amount of substance dissolved in a certain volume of solvent. There are different ways to express this concentration, and they are crucial for accurately describing the composition of a solution. The most common units of solution concentration are molarity (M), molality (m), and mole fraction.

Molarity is defined as the number of moles of solute per liter of solution. Molality, on the other hand, is moles of solute per kilogram of solvent. Unlike molarity, molality is unaffected by changes in temperature and pressure because it's based on the mass of the solvent rather than its volume.

Mole fraction is another way to express concentration, particularly useful in calculations involving partial pressures or colligative properties. The mole fraction of a component in a solution is the ratio of the number of moles of that component to the total number of moles of all components of the solution.
Molality
Molality is an important concentration unit used in chemistry for various calculations, particularly when dealing with temperature or pressure variations. It is defined as the number of moles of solute divided by the kilograms of solvent in the solution.

The formula for calculating molality (m) is given by:\[\begin{equation} \text{Molality} = \frac{\text{Moles of solute}}{\text{Kilograms of solvent}} \end{equation}\]
For instance, if a question states that a glucose solution has a molality of 0.10 m, this means that in every kilogram of water, there are 0.10 moles of glucose dissolved. This unit of measurement is especially useful because it does not change with temperature, making it ideal for studies involving temperature-dependent phenomena such as boiling point elevation or freezing point depression.

Pro tip:

When dealing with molality, always ensure to convert the mass of the solvent to kilograms and use the correct molar mass for the solute for an accurate calculation.

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Most popular questions from this chapter

To prepare a very dilute solution, it is advisable to perform successive dilutions of a single prepared reagent solution, rather than to wcigh a very small mass or to measure a very small volume of stock chemical. A solution was prepared by transferring \(0.661 \mathrm{~g}\) of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) to a \(250.0\)-mL volumetric flask and adding water to the mark. A \(1.000-\mathrm{mL}\). sample of this solution. was transferred to a 500 -mL. volumetric flask and diluted to the mark with water. Then \(10.0 \mathrm{~mL}\) of the diluted solution was transferred to a 250 -mL. flask and diluted to the mark with water. (a) What is the final concentration of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{\text {, in solution? (b) What }}\) mass of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) is in this final solution? (The answer to the last question gives the amount that would have had to have been weighed out if the solution had been prepared directly.)

A chemist studying the propertics of photographic emulsions needed to prepare \(25.00 \mathrm{~mL}\). of \(0.155 \mathrm{M}\) \(\mathrm{AgNO}_{3}\) (aq). What mass of silver nitrate must be placed into a \(25.00-\mathrm{mL}\) volumctric flask and dissolved and diluted to the mark with water?

Describe the preparation of cach solution, starting with the anhydrous solute and warer and using the corrcsponding volumetric flask: (a) \(25.0 \mathrm{~mL}\) of \(6.0 \mathrm{M}\) \(\mathrm{NaOH}(\mathrm{aq}) ;\) (b) \(1.0 \mathrm{~L}\), of \(0.10 \mathrm{M} \mathrm{BaCl}_{2}\) (aq); (c) \(500 \mathrm{~mL}\) of \(0.0010 \mathrm{M} \mathrm{AgNO}_{3}(\mathrm{aq})\).

(a) What volume of a \(0.778 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}\) (aq) solution should be diluted to \(150.0 \mathrm{~mL}\) with water to reduce its concentration to \(0.0234 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{aq})\) ? (b) An experiment requires the use of \(60.0 \mathrm{~mL}\) of \(0.50 \mathrm{M} \mathrm{NaOH}\) (aq). The stockroom assistant can only find a reagent bottle of \(2.5 \mathrm{M} \mathrm{NaOH}\) (aq). How is the \(0.50 \mathrm{M} \mathrm{NaOH}(\) aq \()\) solution to be prepared?

What physical propertics are used for the separation of the components of a mixture by (a) filrration; (b) chromatography; (c) distillation?
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