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Chapter 7: Problem 247

Ethanol boils at \(78.5^{\circ} \mathrm{C}\). If 10 g of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) is dissolved in \(150 \mathrm{~g}\) of ethanol, at what temperature will the solution boil? Assume \(\mathrm{k}_{\mathrm{b}}=\left(1.20^{\circ} \mathrm{C} / \mathrm{M}\right)\) for the alcohol.

Short Answer

Expert verified
The boiling point of the solution is approximately \(78.73^{\circ} \mathrm{C}\) after calculating the molality of the sucrose-ethanol solution and using the boiling point elevation formula.

Step by step solution

01

Calculate the molality of the solution

First, we need to calculate the molality (\(\textit{m}\)) of the solution. Molality is defined as the number of moles of solute per kilogram of solvent. To find the molality, we need to find the moles of sucrose and divide that by the mass of the solvent (ethanol) in kilograms. The molar mass of sucrose (C\(_{12}\)H\(_{22}\)O\(_{11}\)) is: $$ 12 \times 12.01 + 22 \times 1.01 + 11 \times 16.00 = 342.30 \frac{g}{mol} $$ Now, find the moles of sucrose, n\(_\text{sucrose}\), $$ \frac{10 \mathrm{~g}}{342.30\frac{g}{mol}} = 0.0292 \mathrm{~mol} $$ Finally, find the molality, \(\textit{m}\), $$ \frac{0.0292 \mathrm{~mol}}{0.15 \mathrm{~kg}} = 0.1947 \mathrm{~mol / kg} $$
02

Apply the boiling point elevation formula

The formula for boiling point elevation is given by: $$ \Delta T_\text{b} = \textit{m} \times k_\text{b} $$ We already know the molality (\(\textit{m}\)) and the given molal boiling point constant (\(k_\text{b}\)) for ethanol is \(1.20^{\circ} \mathrm{C} / \mathrm{M}\). Calculate the boiling point elevation, \(\Delta T_\text{b}\), $$ \Delta T_\text{b} = 0.1947\, \mathrm{M} \times 1.20\, \frac{^\circ \mathrm{C}}{\mathrm{M}} = 0.2336 ^\circ \mathrm{C} $$
03

Find the boiling point of the solution

Now that we know the boiling point elevation, we can find the boiling point of the solution by adding the boiling point elevation to the boiling point of pure ethanol. $$ T_\text{solution} = T_\text{ethanol} + \Delta T_\text{b} $$ Plug in the values and calculate the boiling point of the solution, $$ T_\text{solution} = 78.5^{\circ} \mathrm{C} + 0.2336 ^\circ \mathrm{C} = 78.7336 ^\circ \mathrm{C} $$ Hence, the solution will boil at approximately \(78.73^{\circ} \mathrm{C}\).

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Key Concepts

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

Molality
Molality is a measure of the concentration of a solute in a solution, specifically in terms of moles of solute per kilogram of solvent. Unlike molarity, which is influenced by temperature changes, molality, represented as \(m\), remains constant as it depends on mass rather than volume. Computing molality involves:
  • Determining the moles of solute. For example, if we dissolve 10 g of sucrose (\(\text{C}_{12}\text{H}_{22}\text{O}_{11}\)) in ethanol, we first calculate the moles of sucrose using its molar mass \(342.30 \frac{\text{g}}{\text{mol}}\).
  • Dividing the moles of the solute by the mass of the solvent in kilograms.
In our exercise, the molality of the sucrose-ethanol solution is found to be \(0.1947 \text{ mol/kg}\), meaning for every kilogram of ethanol, there are \(0.1947\) moles of sucrose.
Molar Mass
Molar mass is the mass of one mole of a given substance. It is expressed in grams per mole (\(\text{g/mol}\)).To determine the molar mass of a compound, sum up the atomic masses of its constituent atoms. For instance, sucrose is composed of carbon (C), hydrogen (H), and oxygen (O) atoms.
  • The molar mass of a single carbon atom is approximately \(12.01 \text{ g/mol}\).
  • Hydrogen's molar mass is roughly \(1.01 \text{ g/mol}\), and oxygen鈥檚 is \(16.00 \text{ g/mol}\).
Thus, the molar mass calculation for sucrose (\(\text{C}_{12}\text{H}_{22}\text{O}_{11}\)) is based on 12 carbon atoms, 22 hydrogen atoms, and 11 oxygen atoms, resulting in \(342.30 \text{ g/mol}\). This number is crucial when calculating moles of sucrose from grams.
Sucrose
Sucrose is a common carbohydrate found in many plants and is often referred to as table sugar. Its chemical formula is \(\text{C}_{12}\text{H}_{22}\text{O}_{11}\). As a non-volatile solute, sucrose can alter the boiling and freezing points of solutions, showcasing colligative properties. When dissolved in a solvent like ethanol as in our example, the sucrose molecules disrupt the regular interaction of ethanol molecules, leading to changes in physical properties of the solution. Understanding sucrose鈥檚 role in colligative properties requires awareness of its contribution to boiling point elevation when dissolved in a solvent.
Ethanol Boiling Point
Ethanol is a volatile, flammable alcohol with a boiling point of \(78.5^{\circ} \text{C}\). When a solute such as sucrose is added to ethanol, the solution's boiling point typically increases鈥攁 process called boiling point elevation. It occurs because the solute particles interfere with the solvent molecules鈥� ability to escape as vapor. In the given problem, the original boiling point of pure ethanol鈥攁 fundamental characteristic鈥攑lays a role in calculating the new boiling point of the ethanol-sucrose solution. After determining the elevation due to added sucrose, the solution boils at a slightly higher temperature of \(78.7336^{\circ} \text{C}\).
Colligative Properties
Colligative properties are those that depend on the number of solute particles in a solution rather than the type. Key colligative properties include boiling point elevation, freezing point depression, vapor pressure lowering, and osmotic pressure.In our example, boiling point elevation is demonstrated when sucrose, a non-electrolyte solute, is added to ethanol. The number of sucrose molecules increases the solution's boiling point compared to pure ethanol. The extent of the boiling point elevation (\(\Delta T_b = m \times k_b\)) depends on:
  • The molality of the solution \((m)\).
  • The solvent's ebullioscopic constant \((k_b)\), which is \(1.20^{\circ} \text{C/M}\) for ethanol.
Such properties are critical in understanding how solutions behave differently from pure solvents.

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Most popular questions from this chapter

Assuming that the density of water is \(.9971\left(\mathrm{~g} / \mathrm{cm}^{3}\right)\) at \(25^{\circ} \mathrm{C}\) and that of ice at \(0^{\circ}\) is \(917\left(\mathrm{~g} / \mathrm{cm}^{3}\right)\), what percent of a water jug at \(25^{\circ} \mathrm{C}\) should be left empty so that, if the water freezes, it will just fill the jug?

If the vapor pressure of \(\mathrm{CC} 1_{4}\) (carbon tetrachloride) is \(.132\) atm at \(23^{\circ} \mathrm{C}\) and \(.526 \mathrm{~atm}\) at \(58^{\circ} \mathrm{C}\), what is the \(\Delta \mathrm{H}^{\prime}\) in this temperature range?

The vapor pressure of benzene at \(75^{\circ} \mathrm{C}\) is 640 torr. \(\mathrm{A}\) solution of \(3.68 \mathrm{~g}\) of a solute in \(53.0 \mathrm{~g}\) benzene has a vapor pressure of 615 torr. Calculate the molecular weight of the solute. (MW of benzene \(=78.0 .)\)

What is the approximate boiling point at standard pressure of a solution prepared by dissolving \(234 \mathrm{~g}\) of \(\mathrm{NaCl}\) in \(500 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O} ?\)

Water at \(30^{\circ} \mathrm{C}\) has a vapor pressure of \(31.82 \mathrm{~mm} \mathrm{Hg}\). When \(25.0 \mathrm{~g}\) of ethylene glycol is added to \(1000 \mathrm{~g}\) of water, the vapor pressure is lowered to \(31.59 \mathrm{~mm} \mathrm{Hg}\). Determine the molecular weight of ethylene glycol.
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