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Chapter 11: Problem 70

What volume of ethylene glycol \(\left(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}\right)\), a nonelectrolyte, must be added to \(15.0 \mathrm{~L}\) water to produce an antifreeze solution with a freezing point \(-25.0^{\circ} \mathrm{C} ?\) What is the boiling point of this solution? (The density of ethylene glycol is \(1.11 \mathrm{~g} / \mathrm{cm}^{3}\), and the density of water is \(1.00 \mathrm{~g} / \mathrm{cm}^{3} .\) )

Short Answer

Expert verified
To create an antifreeze solution with a freezing point of -25.0°C, you must add approximately 181.4 mL of ethylene glycol to 15.0 L of water. The boiling point of this solution is about 106.88°C.

Step by step solution

01

Calculate the molality of the solution

We know that the freezing point depression is related to the molality of the solution through the following formula: ΔTf = Kf × m Where ΔTf is the freezing point depression, Kf is the freezing point depression constant for water, and m is the molality of the solution. The freezing point of pure water is at 0.0°C, and we want the new freezing point to be -25.0°C. Therefore, ΔTf = 25.0°C and Kf = 1.86 °C kg/mol (for water). Now we can calculate the molality (m): m = ΔTf / Kf m = 25.0°C / 1.86 °C kg/mol m = \(13.44 \frac{\text{mol}}{\text{kg}}\)
02

Calculate the mass of ethylene glycol

To find the mass of ethylene glycol, we'll need to convert the 15.0 L water to mass by using the density of water: mass of water = volume × density mass of water = 15.0 L × 1.00 g/mL × 1000 mL/L × 1 kg/1000 g mass of water = 15.0 kg Now we can use the mass of water and the molality we found in Step 1: mass of ethylene glycol = molality × mass of water mass of ethylene glycol = \(13.44 \frac{\text{mol}}{\text{kg}}\) × 15.0 kg mass of ethylene glycol ≈ 201.6 g
03

Calculate the volume of ethylene glycol

We are now able to calculate the volume of ethylene glycol needed. We will use its density to convert mass to volume: volume of ethylene glycol = mass / density volume of ethylene glycol = 201.6 g / (1.11 g/mL) volume of ethylene glycol ≈ 181.4 mL
04

Calculate the boiling point elevation of the solution

Now, let's find the boiling point of the solution. The boiling point elevation formula is: ΔTb = Kb × m Where ΔTb is the boiling point elevation, Kb is the boiling point elevation constant for water, and m is the molality of the solution. For water, Kb = 0.512 °C kg/mol. We can now calculate the boiling point elevation: ΔTb = Kb × m ΔTb = 0.512 °C kg/mol × \(13.44 \frac{\text{mol}}{\text{kg}}\) ΔTb ≈ 6.88°C
05

Calculate the boiling point of the solution

The boiling point of pure water is 100.0°C. To find the boiling point of the solution, we will add the boiling point elevation to the boiling point of pure water: boiling point of solution = boiling point of pure water + ΔTb boiling point of solution = 100.0°C + 6.88°C boiling point of solution ≈ 106.88°C So, it is required to add 181.4 mL of ethylene glycol to 15.0 L of water to produce an antifreeze solution with a freezing point of -25.0°C. The boiling point of this solution is approximately 106.88°C.

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Most popular questions from this chapter

a. Calculate the freezing-point depression and osmotic pressure at \(25^{\circ} \mathrm{C}\) of an aqueous solution containing \(1.0 \mathrm{~g} / \mathrm{L}\) of a protein (molar mass \(=9.0 \times 10^{4} \mathrm{~g} / \mathrm{mol}\) ) if the density of the solution is \(1.0 \mathrm{~g} / \mathrm{cm}^{3}\). b. Considering your answer to part a, which colligative property, freezing- point depression or osmotic pressure, would be better used to determine the molar masses of large molecules? Explain.

In order for sodium chloride to dissolve in water, a small amount of energy must be added during solution formation. This is not energetically favorable. Why is \(\mathrm{NaCl}\) so soluble in water?

Rationalize the trend in water solubility for the following simple alcohols:

What mass of glycerin \(\left(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{3}\right)\), a nonelectrolyte, must be dissolved in \(200.0 \mathrm{~g}\) water to give a solution with a freezing point of \(-1.50^{\circ} \mathrm{C} ?\)

A \(1.60-\mathrm{g}\) sample of a mixture of naphthalene \(\left(\mathrm{C}_{10} \mathrm{H}_{8}\right)\) and anthracene \(\left(\mathrm{C}_{14} \mathrm{H}_{10}\right)\) is dissolved in \(20.0 \mathrm{~g}\) benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\). The freezing point of the solution is \(2.81^{\circ} \mathrm{C}\). What is the composition as mass percent of the sample mixture? The freezing point of benzene is \(5.51^{\circ} \mathrm{C}\) and \(K_{\mathrm{f}}\) is \(5.12^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol}\).
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