Question

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 is found to be \(-0.426^{\circ} \mathrm{C}\). Assuming ideal behavior, calculate the mass percent composition of the original mixture, and the mole fraction of sucrose in the original mixture.

Short answer

The mass percent composition of the original mixture is approximately 68.31% sucrose and 31.69% NaCl; and the mole fraction of sucrose in the original mixture is approximately 0.551.

Step-by-step solution

Determine the molality of the solution using the freezing point depression formula

We are given the freezing point depression, ΔTf, as -0.426°C. We also know the freezing point depression constant for water, Kf, is 1.86 °C·kg/mol. Using the freezing point depression formula, we can find the molality: ΔTf = Kf·m -0.426°C = 1.86 °C·kg/mol ·m m = -0.426°C / 1.86 °C·kg/mol ≈ 0.229 mol/kg

Set up equations for moles of sucrose and NaCl

Let S be the moles of sucrose and N be the moles of NaCl. The total mass of the solution is 20.0 g and the mass of water is 1.00 kg. We can set up the following equations: Molality of sucrose: m(S) = S / 1.00 kg Molality of NaCl: m(N) = N / 1.00 kg Total molality: m(S) + m(N) = 0.229 mol/kg Mass of sucrose: S × 342.30 g/mol (molar mass of sucrose) Mass of NaCl: N × 58.44 g/mol (molar mass of NaCl) Total mass: S × 342.30 g/mol + N × 58.44 g/mol = 20.0 g

Solve the system of equations

Use m(S) + m(N) = 0.229 mol/kg and S × 342.30 g/mol + N × 58.44 g/mol = 20.0 g to solve for moles of sucrose (S) and NaCl (N). S + 5.855N ≈ 0.229 (Dividing both sides of the first equation by the molar mass ratio 58.44/342.30 ≈ 5.855) S × 342.30 + N × 58.44 = 20.0 Using substitution or any other suitable method, we get: S ≈ 0.0398 moles N ≈ 0.0324 moles

Calculate the mass percent composition

Mass percent of sucrose: %mass(S) = (S × 342.30 g/mol) / 20.0 g × 100 %mass(S) = (0.0398 mol × 342.30 g/mol) / 20.0 g × 100 ≈ 68.31% Mass percent of NaCl: %mass(N) = (N × 58.44 g/mol) / 20.0 g × 100 %mass(N) = (0.0324 mol × 58.44 g/mol) / 20.0 g × 100 ≈ 31.69%

Calculate the mole fraction of sucrose

Total moles of solute: S + N Total moles = 0.0398 mol + 0.0324 mol ≈ 0.0722 mol Mole fraction of sucrose: X(S) = S / (S + N) X(S) = 0.0398 mol / 0.0722 mol ≈ 0.551 The mass percent composition of the original mixture is approximately 68.31% sucrose and 31.69% NaCl; and the mole fraction of sucrose in the original mixture is approximately 0.551.

Key concepts

Molality

Molality is a way to express the concentration of a solution. It focuses on the amount of solute per kilogram of solvent, unlike molarity, which considers the total volume of the solution. In our example, we used molality to analyze how much solute (sucrose and NaCl) was dissolved in 1 kg of water.

To calculate molality, the equation is simple:

• \[ m = \frac{\text{moles of solute}}{\text{kilograms of solvent}} \]

In the exercise, we found the molality by using the freezing point depression formula, a property that helps us understand how solutes affect the freezing point of a solvent like water.

Molality is particularly useful in studies involving temperature changes because it's independent of temperature, giving it an advantage over some other concentration units.

Mass Percent Composition

Mass percent composition tells us how much of each component is present in a mixture by mass. Think of it as calculating what percentage of a mixture's mass comes from each component. In this calculation, the mass of each substance is essential.

For our sucrose and NaCl mixture, we found:

• Mass percent of sucrose: \[ \frac{\text{mass of sucrose}}{\text{total mass of mixture}} \times 100 \]
• Mass percent of NaCl: \[ \frac{\text{mass of NaCl}}{\text{total mass of mixture}} \times 100 \]

By plugging in the masses, we discovered that the sucrose formed about 68.31% of the mixture, and NaCl accounted for about 31.69%.

This method helps chemists and scientists quickly grasp the relative abundance of substances in a mixture, which can be crucial for both reactions and production processes.

Mole Fraction

The mole fraction is a way to express the concentration of a component in a mixture by looking at its proportion in relation to the total number of moles. This dimensionless quantity helps to understand the composition of mixtures on a mole basis.

In this context, the mole fraction of sucrose was calculated using:

• \[ X(S) = \frac{\text{moles of sucrose}}{\text{total moles of solute}} \]

The calculation showed that sucrose made up approximately 0.551 of the total moles in the solution.

Understanding mole fractions is fundamental in chemistry because it offers a clear picture of how a mixture is composed, which is especially useful in predicting and explaining how solutions will behave in chemical reactions or physical processes.