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Chapter 13: Problem 30

A \(10.00 \%\) -by-mass solution of ethanol, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) in water has a density of \(0.9831 \mathrm{g} / \mathrm{mL}\) at \(15^{\circ} \mathrm{C}\) and \(0.9804 \mathrm{g} / \mathrm{mL}\) at \(25^{\circ} \mathrm{C} .\) Calculate the molality of the ethanol-water solution at these two temperatures. Does the molality differ at the two temperatures (that is, at 15 and \(25^{\circ} \mathrm{C}\) )? Would you expect the molarities to differ? Explain.

Short Answer

Expert verified
The calculated molality of the ethanol-water solution is 2.41 moles/kg at both 15°C and 25°C. Despite the temperature change, the molality remains the same as it only depends on the mass of the solvent (water in this case) and the number of moles of solute (ethanol), and these do not change with temperature. On the other hand, the molarities would be expected to differ at the two temperatures, since molarity is dependent on the volume of the solution, and the volume can change with temperature.

Step by step solution

01

Understanding Molality

Molality \(m\) is a measure of the concentration of a solute in a solution, or a measure of the moles of solute per kilogram of solvent. Its unit is moles/kg.
02

Calculating the Mass of the Solution

In this case, using the mass percentage and density at 15°C, the mass of the solution can be calculated. The density given (0.9831 g/mL) is equivalent to 0.9831 kg/L. Since the solution is 10% ethanol by mass, the mass of ethanol in 1 L of solution is 0.10 * 0.9831 kg = 0.09831 kg.
03

Calculating Molality at 15°C

Once the mass of the ethanol is known, the molality can be calculated. The molar mass of ethanol \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) is approximately 46.07 g/mol or 0.04607 kg/mol. The number of moles of ethanol in the solution is therefore 0.09831 kg / 0.04607 kg/mol = 2.133 moles. The mass of the water in the solution is 0.9831 kg - 0.09831 kg = 0.88479 kg. Thus, the molality of the solution at 15°C is 2.133 moles / 0.88479 kg = 2.41 moles/kg.
04

Repeating Calculations for 25°C

The above calculations can be repeated for the solution at 25°C with a density of 0.9804 kg/L. The mass of ethanol in the solution would be 0.10 * 0.9804 kg = 0.09804 kg and the mass of the water would be 0.9804 kg - 0.09804 kg = 0.88236 kg. Thus, the molality at 25°C is 2.13 moles / 0.88236 kg = 2.41 moles/kg.
05

Comparing the Molalities

Observing the molality at both temperatures, it can be seen that they are equal - the molality does not change with temperature.
06

Discussing Molarity

However, molarity - moles of solute per litre of solution - would be expected to differ at different temperatures. This is because molarity depends on the volume of the solution, which can change with temperature due to thermal expansion or contraction.

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Most popular questions from this chapter

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.

At \(25^{\circ} \mathrm{C}\) a \(0.50 \mathrm{g}\) sample of polyisobutylene (a polymer used in synthetic rubber) in \(100.0 \mathrm{mL}\) of benzene solution has an osmotic pressure that supports a \(5.1 \mathrm{mm}\) column of solution \((d=0.88 \mathrm{g} / \mathrm{mL}) .\) What is the molar mass of the polyisobutylene? (For \(\mathrm{Hg}\), \(d=13.6 \mathrm{g} / \mathrm{mL} .)\)

The solubility of a nonreactive gas in water increases with (a) an increase in gas pressure; (b) an increase in temperature; (c) increases in both temperature and pressure; (d) an increase in the volume of gas in equilibrium with the available water.

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