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Chapter 13: Problem 71

Adding \(1.00 \mathrm{g}\) of benzene, \(\mathrm{C}_{6} \mathrm{H}_{6},\) to \(80.00 \mathrm{g}\) cyclohexane, \(\mathrm{C}_{6} \mathrm{H}_{12},\) lowers the freezing point of the cyclohexane from 6.5 to \(3.3^{\circ} \mathrm{C}\). (a) What is the value of \(K_{f}\) for cyclohexane? (b) Which is the better solvent for molar mass determinations by freezing- point depression, benzene or cyclohexane? Explain.

Short Answer

Expert verified
The value of \(K_{f}\) for cyclohexane is 20°C kg/mol. The solvent with the larger value of \(K_{f}\) would be a better solvent for molar mass determinations by freezing-point depression. Without the value of \(K_{f}\) for benzene, it can't be definitively said which solvent is better.

Step by step solution

01

Calculate the ΔTf

The depression in freezing point, ΔTf, is equal to the original freezing point of cyclohexane minus the new freezing point after the benzene has been added. Hence, ΔTf = 6.5°C - 3.3°C = 3.2°C.
02

Convert grams of benzene to moles

On the other hand, the molecular weight of benzene, \(\mathrm{C}_{6} \mathrm{H}_{6}\), is 78.11 g/mol. Therefore, to get the number of moles, divide the given mass of benzene by the molecular weight. The result is 1.00g ÷ 78.11g/mol = 0.0128 mol benzene.
03

Convert grams of cyclohexane to kilograms

We then convert the mass of the cyclohexane solvent to kilograms since molality is represented in terms of kilogram. Hence, 80.00g of cyclohexane = 80.00g ÷ 1000 = 0.0800 kg.
04

Calculate the molality (m)

Having obtained the values of moles of solute and mass of solvent in kilograms, the molality can be calculated as: \( m = \frac{moles \, of \, solute}{mass \, of \, solvent \, (kg)} \) = \(\frac {0.0128}{0.0800}\) = 0.160 mol/kg.
05

Calculate Kf

With the freezing point depression ΔTf = 3.2°C and the molality m = 0.16 mol/kg, the depression constant can be calculated using the formula ΔTf = \(K_f \times m\). Rearranging, \(K_f = \frac {ΔTf}{m}\) =\(\frac{3.2}{0.16}\) = 20°C kg/mol.
06

Determine which is the better solvent

For molar mass determinations by freezing point depression, a better solvent would have a larger \(\(K_f\)\) value because a larger \(\(K_f\)\) would yield a larger change in freezing temperature for a given amount of solute. This makes it easier to measure the change in freezing point. Therefore, whichever between benzene and cyclohexane has the higher \(\(K_f\)\) will be considered the better solvent.

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Most popular questions from this chapter

Calculate the mole fraction of the solute in the following aqueous solutions:(a) \(0.112 \mathrm{M} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) \((d=1.006 \mathrm{g} / \mathrm{mL})\) (b) \(3.20 \%\) ethanol,by volume \((d=0.993 \mathrm{g} / \mathrm{mL}\) pure \(\left.\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}, d=0.789 \mathrm{g} / \mathrm{mL}\right)\).

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