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Chapter 13: Problem 41

A solution is made containing \(50.0 \mathrm{~g}\) of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) in \(1000 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O} .\) Calculate \((\mathbf{a})\) the mole fraction of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH},\) (b) the mass percent of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH},(\mathbf{c})\) the molality of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\).

Short Answer

Expert verified
Mole fraction: 0.0193, Mass percent: 4.76%, Molality: 1.085 mol/kg.

Step by step solution

01

Calculate the Moles of Ethanol

First, find the molar mass of ethanol (\(\mathrm{C}_2\mathrm{H}_5\mathrm{OH}\)). Ethanol has 2 carbon atoms, 6 hydrogen atoms, and 1 oxygen atom. Therefore, its molar mass is calculated as:\[\left(2 \times 12.01\right) + \left(6 \times 1.008\right) + \left(1 \times 16.00\right) = 46.08 \, \text{g/mol}\]. Now, calculate the number of moles of ethanol using the formula: \(\text{moles} = \frac{\text{mass}}{\text{molar mass}}\). So, \[\text{moles of } \mathrm{C}_2\mathrm{H}_5\mathrm{OH} = \frac{50.0 \text{ g}}{46.08 \text{ g/mol}} \approx 1.085 \text{ moles}\].
02

Calculate the Mole Fraction of Ethanol

Find the moles of water (\(\mathrm{H}_2\mathrm{O}\)). The molar mass of water is \(18.015\, \text{g/mol}\). Thus, \[\text{moles of } \mathrm{H}_2\mathrm{O}= \frac{1000 \text{ g}}{18.015 \text{ g/mol}} \approx 55.51 \text{ moles}\]. Now, use the mole fraction formula: \(\chi = \frac{\text{moles of solute}}{\text{moles of solute} + \text{moles of solvent}}\). Substitute the known values: \[\chi_{\mathrm{C}_2\mathrm{H}_5\mathrm{OH}} = \frac{1.085}{1.085 + 55.51} \approx 0.0193\].
03

Calculate the Mass Percent of Ethanol

Mass percent is calculated by the formula \(\text{mass percent} = \left(\frac{\text{mass of solute}}{\text{mass of solution}} \right) \times 100\%\). The mass of the solution is \(50.0 \text{ g (ethanol)} + 1000 \text{ g (water)} = 1050 \text{ g}\). Therefore, the mass percent of ethanol is: \[\frac{50.0}{1050} \times 100\% \approx 4.76\%\].
04

Calculate the Molality of Ethanol

Molality is defined as the moles of solute per kilogram of solvent. Using the moles for ethanol calculated earlier (1.085 moles) and the mass of the solvent (water) in kilograms, which is 1.000 kg, compute: \[\text{molality} = \frac{1.085}{1.000} = 1.085 \text{ mol/kg}\].

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Key Concepts

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

Mole Fraction
The mole fraction is a way to express the concentration of a component in a mixture. It's the ratio of the number of moles of one component to the total number of moles of all components in the solution. In this exercise, the mole fraction of ethanol was calculated by dividing the moles of ethanol by the sum of moles of ethanol and water. This provides a dimensionless number, which indicates the proportion of ethanol molecules among all molecules in the solution.
  • Number of moles of ethanol: Calculated using its mass and molar mass.
  • Number of moles of water: Found by dividing the mass of water by its molar mass.
  • Mole fraction formula: \[ \chi = \frac{\text{moles of solute}}{\text{moles of solute} + \text{moles of solvent}} \]
  • Result: 0.0193, indicating ethanol is a minor component relative to water.

Knowing the mole fraction helps in understanding the composition and predicting the behavior of solutions, especially in physical chemistry.
Mass Percent
Mass percent is another way to describe the concentration of a substance in a solution. It tells you what fraction of the total mass of the solution is made up of a particular solute. This is useful for practical tasks like preparing a solution to a specific concentration.
  • The formula used is: \[ \text{Mass percent} = \left(\frac{\text{mass of solute}}{\text{mass of solution}} \right) \times 100\% \]
  • In this exercise, the mass of the solution is the sum of the mass of ethanol and water.
  • The calculated mass percent for ethanol is approximately 4.76%.

Mass percent is commonly used in laboratory settings for creating solutions with precise amounts of solute. It's a straightforward way to communicate how much of a substance is contained in a solution.
Molality
Molality is an important concentration unit in solution chemistry, especially when dealing with temperature-dependent scenarios. Unlike molarity, molality is independent of temperature since it is based on the mass of the solvent, not its volume.
  • Calculated using: \[ \text{Molality} = \frac{\text{moles of solute}}{\text{kilograms of solvent}} \]
  • For our solution, the number of moles of ethanol was divided by the mass of water in kilograms.
  • The result was a molality of 1.085 mol/kg.


Molality is particularly useful when the solution undergoes temperature changes, as the solvent's volume and solute's concentration might vary, but the mass remains constant. Thus, it's often preferred in thermodynamic calculations.

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Most popular questions from this chapter

Calculate the number of moles of solute present in each of the following aqueous solutions: (a) \(750 \mathrm{~mL}\) of \(0.120 \mathrm{M}\) \(\operatorname{SrBr}_{2},(\mathbf{b}) 70.0 \mathrm{~g}\) of \(0.200 \mathrm{~m} \mathrm{KCl},(\mathbf{c}) 150.0 \mathrm{~g}\) of a solution that is \(5.75 \%\) glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) by mass.

\(\mathrm{KBr}\) is relatively soluble in water, yet its enthalpy of solution is \(+19.8 \mathrm{~kJ} / \mathrm{mol}\). Which of the following statements provides the best explanation for this behavior? (a) Potassium salts are always soluble in water. (b) The entropy of mixing must be unfavorable. (c) The enthalpy of mixing must be small compared to the enthalpies for breaking up water-water interactions and K-Br ionic interactions. (d) \(\mathrm{KBr}\) has a high molar mass compared to other salts like \(\mathrm{NaCl}\)

An ionic compound has a very negative \(\Delta H_{\text {soln }}\) in water. (a) Would you expect it to be very soluble or nearly insoluble in water? (b) Which term would you expect to be the largest negative number: \(\Delta H_{\text {solvent }}, \Delta H_{\text {solute }}\), or \(\Delta H_{\operatorname{mix}} ?\)

The vapor pressure of pure water at \(70^{\circ} \mathrm{C}\) is \(31.2 \mathrm{kPa}\). The vapor pressure of water over a solution at \(70^{\circ} \mathrm{C}\) containing equal numbers of moles of water and glycerol \(\left(\mathrm{C}_{3} \mathrm{H}_{5}(\mathrm{OH})_{3}\right.\), a nonvolatile solute) is \(13.3 \mathrm{kPa}\). Is the solution ideal according to Raoult's law?

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.
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