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Chapter 12: Problem 33

A \(10.0 \%\) solution of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) in water has a density of \(1.038 \mathrm{~g} / \mathrm{mL}\). Express the concentration of the sugar as (a) molality. (b) molarity. (c) mole fraction.

Short Answer

Expert verified
Molality is 0.324 mol/kg, molarity is 0.303 M, and mole fraction is 0.0058.

Step by step solution

01

Understand the Problem

We have a 10.0% sucrose solution, which means 10g of sucrose per 100g of the solution. The density of the solution is 1.038 g/mL. We need to find molality, molarity, and mole fraction of sucrose.
02

Convert Mass to Moles

First, calculate the moles of sucrose in 10 g. The molar mass of sucrose is \(C_{12}H_{22}O_{11} = 12 \times 12 + 22 \times 1 + 11 \times 16 = 342 \text{ g/mol}\). Thus, the moles of sucrose are \(\frac{10 \text{ g}}{342 \text{ g/mol}} = 0.0292 \text{ mol}.\)
03

Calculate Molality

Molality (m) is moles of solute per kg of solvent. We have 10 g of sucrose in 100 g of solution, implying \(100 - 10 = 90 \text{ g}\) (0.090 kg) of water. \[ \text{Molality} = \frac{0.0292 \text{ mol}}{0.090 \text{ kg}} = 0.324 \text{ mol/kg}. \]
04

Calculate Volume of Solution

Using the density (1.038 g/mL), the volume of 100 g of solution is \(\frac{100 \text{ g}}{1.038 \text{ g/mL}} \approx 96.34 \text{ mL} = 0.0963 \text{ L}.\)
05

Calculate Molarity

Molarity (M) is moles of solute per liter of solution.\[ \text{Molarity} = \frac{0.0292 \text{ mol}}{0.0963 \text{ L}} \approx 0.303 \text{ M}. \]
06

Calculate Mole Fraction of Sucrose

Mole fraction \(X\) of sucrose is the ratio of moles of sucrose to total moles in the solution. We already have 0.0292 moles of sucrose. The moles of water are \(\frac{90 \text{ g}}{18 \text{ g/mol}} = 5 \text{ mol}.\) Thus, \[ X_{\text{sucrose}} = \frac{0.0292}{0.0292 + 5} = 0.0058. \]

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Key Concepts

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

Molality
Molality is a way to express the concentration of a solution. It specifically measures the amount of solute in relation to the mass of the solvent, which in many cases is water. The formula for molality is defined as the number of moles of solute per kilogram of solvent. This measurement is particularly useful because, unlike molarity, it is not affected by temperature or pressure.
Here's how to determine molality:
  • First, find the moles of the solute. Using sucrose as an example, we have already calculated that there are 0.0292 moles in 10 grams of sucrose.
  • Next, convert the mass of the solvent from grams to kilograms. In our example, if the total weight of the solution is 100 grams and 10 grams is sucrose, then 90 grams is the solvent (water). This is 0.090 kilograms.
  • The formula is then: \[ \text{Molality} (m) = \frac{0.0292 \text{ mol}}{0.090 \text{ kg}} = 0.324 \text{ mol/kg}. \]
Molality provides a consistent way to describe concentration regardless of external conditions.
Molarity
Molarity is another term used to define the concentration of a solution, which tells us how many moles of a solute are present in one liter of the solution. It is one of the most commonly used units of concentration in chemistry due to its straightforward relationship with volume.
To determine molarity, follow these steps:
  • Calculate the moles of solute present. We have seen that there are 0.0292 moles of sucrose in our example.
  • Then, use the density of the solution to find its volume. Given the density of 1.038 g/mL, the total solution mass of 100g equates to a volume of approximately 96.34 mL or 0.0963 L.
  • The molarity calculation is: \[ \text{Molarity} (M) = \frac{0.0292 \text{ mol}}{0.0963 \text{ L}} \approx 0.303 \text{ M}. \]
Bear in mind, molarity is impacted by changes in temperature, as it involves the volume that can expand or contract.
Mole Fraction
The mole fraction is a dimensionless quantity that expresses the ratio of the number of moles of one component to the total number of moles in the mixture. It provides detailed insight into the ratio of components in a solution.
To calculate the mole fraction:
  • Identify the moles of the solute, which for sucrose, we determined to be 0.0292 moles.
  • Assume the solution includes water as the other component. The moles of water are found by dividing the mass of the water (90 g) by its molar mass (18 g/mol), equating to 5 moles.
  • The mole fraction of sucrose is calculated as: \[ X_{\text{sucrose}} = \frac{0.0292}{0.0292 + 5} = 0.0058. \]
Mole fractions are often used in thermodynamics and colligative properties, where the precise relationship between components is important.

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Most popular questions from this chapter

A solution is prepared by dissolving \(25.0 \mathrm{~g} \mathrm{BaCl}_{2}\) in \(500 \mathrm{~g}\) water. Express the concentration of \(\mathrm{BaCl}_{2}\) in this solution as (a) mass percentage. (b) mole fraction. (c) molality.

Choose the solute of each pair that would be more soluble in water. Explain your answer. (a) \(\mathrm{NaOH}\) or \(\mathrm{CO}_{2}\) (b) \(\mathrm{TiCl}_{3}\) or \(\mathrm{CHCl}_{3}\) (c) \(\mathrm{C}_{3} \mathrm{H}_{8}\) or \(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{OH}\)

Choose the solute of each pair that would be more soluble in hexane \(\left(\mathrm{C}_{6} \mathrm{H}_{14}\right)\). Explain your answer. (a) \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{10} \mathrm{OH}\) or \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{2} \mathrm{OH}\) (b) \(\mathrm{BaCl}_{2}\) or \(\mathrm{CCl}_{4}\) (c) \(\mathrm{Fe}\left(\mathrm{C}_{5} \mathrm{H}_{5}\right)_{2}\) ( a nonelectrolyte) or \(\mathrm{FeCl}_{2}\)

Suppose you have two aqueous solutions separated by a semipermeable membrane. One contains \(5.85 \mathrm{~g} \mathrm{NaCl}\) dissolved in \(100 \mathrm{~mL}\) of solution, and the other contains \(8.88 \mathrm{~g} \mathrm{KNO}_{3}\) dissolved in \(100 \mathrm{~mL}\) of solution. In which direction will solvent flow: from the \(\mathrm{NaCl}\) solution to the \(\mathrm{KNO}_{3}\) solution, or from \(\mathrm{KNO}_{3}\) to \(\mathrm{NaCl}\) ? Explain briefly.

We analyzed the enthalpy change that accompanies the formation of a solution from the pure solute and solvent by considering three changes that must occur: (1) separate the solvent molecules; (2) separate the solute molecules; and (3) bring the solute particles and solvent particles together. If the enthalpy of solution is negative (an exothermic process), what can be said about the relative enthalpy changes of the three processes just listed?
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