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Chapter 11: Problem 109

You make \(20.0 \mathrm{~g}\) of a sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) and \(\mathrm{NaCl}\) mixture and dissolve it in \(1.00 \mathrm{~kg}\) water. The freezing point of this solution s found to be \(-0.426^{\circ} \mathrm{C}\). Assuming ideal behavior, calculate the nass percent composition of the original mixture, and the mole fraction of sucrose in the original mixture.

Short Answer

Expert verified
The mass percent composition of the original mixture is 71.8 % sucrose and 28.2 % NaCl. The mole fraction of sucrose in the original mixture is 0.303.

Step by step solution

01

Determine the molality of the solution

To determine the molality of the solution, we will use the freezing point depression formula: 螖T = Kf * molality Here, 螖T represents the change in freezing point, Kf is the cryoscopic constant of water (1.86 掳Ckg/mol), and molality is the molality of the solution in mol/kg. From the given data, 螖T = 0.426 掳C, so we have: 0.426 掳C = 1.86 掳Ckg/mol * molality Solving for molality, we get: molality = 0.426 掳C / 1.86 掳Ckg/mol = 0.229 mol/kg
02

Calculate the moles of sucrose and NaCl

Let x be the mass of sucrose. Then, the mass of NaCl would be (20-x g). Since molality represents the ratio of moles of solute to the mass of the solvent, we can write the molality of the solution as follows: molality = (moles of sucrose + moles of NaCl) / 1 kg Using the molar masses of sucrose (\(M = 342.30 \mathrm{~g/mol}\)) and NaCl (\(M = 58.44 \mathrm{~g/mol}\)), we can rewrite the equation as: 0.229 mol/kg = ((x g) / 342.30 g/mol + ((20 - x) g) / 58.44 g/mol) / 1 kg
03

Solve for the mass of sucrose

Now, we can solve the equation above to find the mass of sucrose (x): 0.229 mol/kg = (x g / 342.30 g/mol + (20 - x) g / 58.44 g/mol) / 1 kg Multiplying both sides by the denominators, we get: 0.229 mol/kg * (342.30 g/mol * 58.44 g/mol) = x g * 58.44 g/mol + (20 - x) g * 342.30 g/mol Solving for x, we find: x = 14.36 g
04

Calculate the mass percent composition and mole fraction

Using the mass of sucrose and the mixture, we can find the mass percent composition of sucrose and NaCl: Mass percent of sucrose = (14.36 g / 20 g) * 100 = 71.8 % Mass percent of NaCl = 100 % - 71.8 % = 28.2 % To find the mole fraction, we first need to find the moles of sucrose and NaCl: Moles of sucrose = 14.36 g / 342.30 g/mol = 0.0419 mol Moles of NaCl = (20 g - 14.36 g) / 58.44 g/mol = 0.0963 mol Now, we can calculate the mole fraction of sucrose: Mole fraction of sucrose = moles of sucrose / (moles of sucrose + moles of NaCl) = 0.0419 mol / (0.0419 mol + 0.0963 mol) = 0.303 So, the mass percent composition of the original mixture is 71.8 % sucrose and 28.2 % NaCl, and the mole fraction of sucrose in the original mixture is 0.303.

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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 solvent, defined as the number of moles of solute divided by the mass of the solvent in kilograms. Unlike molarity, molality is not affected by temperature changes because it is based on mass, not volume.

In the context of freezing point depression, molality plays a crucial role as it directly correlates to how much the freezing point of a solution is lowered compared to the pure solvent. A higher molality means more solute particles are present, which interfere with the formation of a solid phase, causing a greater freezing point depression. When solving problems related to freezing point depression, understanding how to calculate molality is essential, just as we determined the molality using the formula \(\Delta T = K_f \times \text{molality}\) in our example.
Sucrose and NaCl Solution
When dissolving a mixture like sucrose (\(\mathrm{C}_{12}\mathrm{H}_{22}\mathrm{O}_{11}\)) and sodium chloride (NaCl) in water, a solution is formed that combines the properties of both solutes.

Sucrose is a non-electrolyte, meaning it does not dissociate into ions in a solution, whereas NaCl is a strong electrolyte and dissociates completely. This difference is important in freezing point depression, as the number of particles affects the overall effect. While sucrose contributes its molecules to the solution, NaCl contributes both sodium and chloride ions, effectively doubling the number of particles and thus having a potentially greater impact on the freezing point than an equal number of sucrose molecules.
Mass Percent Composition
Mass percent composition refers to the percentage of a component's mass related to the total mass of the mixture. It is a way of expressing the concentration of an ingredient in a mixture.

To find the mass percent composition, you divide the mass of the component of interest by the total mass of the mixture and then multiply the result by 100%. Calculations of mass percent composition are widely used in chemistry for various applications including the analysis of compounds and solutions. In our exercise, we calculated the mass percent composition of both sucrose and NaCl, allowing us to understand the distribution of these components in the original mixture.
Mole Fraction Calculation
The mole fraction is a dimensionless quantity that represents the ratio of the number of moles of a given component to the total number of moles of all components in the mixture. It is an important concept when working with mixtures and solutions as it provides a measure of the composition without the need for a specific volume or mass.

To find the mole fraction, you divide the numbers of moles of the component by the total moles in the mixture. The mole fraction is useful in various chemical calculations, such as determining vapor pressure and the boiling point elevation. As in our case study, calculating the mole fraction of sucrose helped us understand its proportion in relation to NaCl in the mixture.

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Most popular questions from this chapter

In order for sodium chloride to dissolve in water, a small amount of energy must be added during solution formation. This is not energetically favorable. Why is \(\mathrm{NaCl}\) so soluble in water?

Plants that thrive in salt water must have internal solutions (inside the plant cells) that are isotonic with (have the same osmotic pressure as) the surrounding solution. A leaf of a saltwater plant is able to thrive in an aqueous salt solution (at \(\left.25^{\circ} \mathrm{C}\right)\) that has a freezing point equal to \(-0.621^{\circ} \mathrm{C}\). You would like to use this information to calculate the osmotic pressure of the solution in the cell. a. In order to use the freezing-point depression to calculate osmotic pressure, what assumption must you make (in addition to ideal behavior of the solutions, which we will assume)? b. Under what conditions is the assumption (in part a) reasonable? c. Solve for the osmotic pressure (at \(25^{\circ} \mathrm{C}\) ) of the solution in the plant cell. d. The plant leaf is placed in an aqueous salt solution (at \(\left.25^{\circ} \mathrm{C}\right)\) that has a boiling point of \(102.0^{\circ} \mathrm{C}\). What will happen to the plant cells in the leaf?

Which solvent, water or carbon tetrachloride, would you choose to dissolve each of the following? a. \(\mathrm{KrF}_{2}\) e. \(\mathrm{MgF}_{2}\) b. \(\mathrm{SF}_{2}\) f. \(\mathrm{CH}_{2} \mathrm{O}\) c. \(\mathrm{SO}_{2}\) g. \(\mathrm{CH}_{2}=\mathrm{CH}_{2}\) d. \(\mathrm{CO}_{2}\)

Calculate the solubility of \(\mathrm{O}_{2}\) in water at a partial pressure of \(\mathrm{O}_{2}\) of 120 torr at \(25^{\circ} \mathrm{C}\). The Henry's law constant for \(\mathrm{O}_{2}\) is \(1.3 \mathrm{X}\) \(10^{-3} \mathrm{~mol} / \mathrm{L} \cdot\) atm for Henry's law in the form \(C=k P\), where \(C\) is the gas concentration \((\mathrm{mol} / \mathrm{L})\).

a. Calculate the freezing-point depression and osmotic pressure at \(25^{\circ} \mathrm{C}\) of an aqueous solution containing \(1.0 \mathrm{~g} / \mathrm{L}\) of a protein (molar mass \(=9.0 \times 10^{4} \mathrm{~g} / \mathrm{mol}\) ) if the density of the solution is \(1.0 \mathrm{~g} / \mathrm{cm}^{3}\). b. Considering your answer to part a, which colligative property, freezing- point depression or osmotic pressure, would be better used to determine the molar masses of large molecules? Explain.
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