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Chapter 11: Problem 129

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

Expert verified
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

01

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
02

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
03

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
04

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%
05

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.

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