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Chapter 10: Problem 67

Pure benzene boils at \(80.10^{\circ} \mathrm{C}\) and has a boiling point constant, \(k_{\mathrm{b}}\), of \(2.53^{\circ} \mathrm{C} / \mathrm{m} .\) A sample of benzene is contaminated by naphthalene, \(\mathrm{C}_{10} \mathrm{H}_{8}\). The boiling point of the contaminated sample is \(81.20^{\circ} \mathrm{C}\). How pure is the sample? (Express your answer as mass percent of benzene.)

Short Answer

Expert verified
The mass percent of benzene in the contaminated sample is approximately 91.88%.

Step by step solution

01

Calculate the change in boiling point (ΔT)

To find the change in boiling point, we can subtract the boiling point of pure benzene from the boiling point of the contaminated sample: ΔT = \(81.20^\circ C - 80.10^\circ C = 1.10^\circ C\)
02

Calculate the molality of the solution

To find the molality, use the boiling point elevation formula and solve for molality: molality = ΔT / \(k_b\) molality = \(1.10^\circ C / 2.53^\circ C/ \mathrm{m} = 0.435 \mathrm{m}\)
03

Molecular weights of benzene and naphthalene

The molecular weight of benzene (C6H6) is \(6(12.01) + 6(1.01) = 78.12 \ g/mol\). The molecular weight of naphthalene (C10H8) is \(10(12.01) + 8(1.01) = 128.18 \ g/mol\).
04

Let x be the mass naphthalene in grams

We have the molality of the solution (0.435). Molality is defined as moles of solute (naphthalene) per kilogram of solvent (benzene). If we let x be the mass of naphthalene in grams, we have: molality = moles of naphthalene / mass of benzene in kg 0.435 = \([(x \ g) / (128.18 \ g/mol)] / [(1000 - x) \ g * (1 \ kg / 1000 \ g)]\)
05

Solve for x - mass of naphthalene

Multiply both sides by (1000 - x) and simplify: 0.435(1000 - x) = (x) / (128.18) 435 - 0.435x = x / 128.18 0.435x + x / 128.18 = 435 x(0.435 + 1 / 128.18) = 435 x = 435 / (0.435 + 1 / 128.18) x = 81.16
06

Calculate the mass percent of benzene

Finally, we can find the mass percent of benzene in the contaminated sample as follows: Mass percent of benzene = (mass of benzene / total mass) * 100 = \([(1000 - 81.16) \ g / 1000 \ g] * 100\) = \(918.84/1000 * 100\) = \(91.88 \%\) The contaminated sample of benzene is approximately \(91.88\%\) pure.

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