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Chapter 13: Problem 75

How many grams of ethylene glycol \(\left(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}\right)\) must be added to \(1.00 \mathrm{~kg}\) of water to produce a solution that freezes at \(-5.00^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
Approximately 166.96 grams of ethylene glycol (C2H6O2) must be added to 1.00 kg of water to produce a solution that freezes at -5.00掳C.

Step by step solution

01

Calculate the freezing-point depression

The freezing-point depression, 螖Tf, is the difference between the freezing point of the pure solvent and the freezing point of the solution. In this case, 螖Tf is the difference between the freezing point of pure water, 0掳C, and the freezing point of the solution, -5.00掳C. 螖Tf = 0 - (-5.00) 螖Tf = 5.00掳C
02

Use the freezing-point depression formula to find the molality of the solution

The freezing-point depression can be calculated using the formula: 螖Tf = Kf 鈰� molality Where Kf is the cryoscopic constant of a solvent (in the case of water, Kf = 1.86 掳C/molal), and molality is the number of moles of solute per kilogram of solvent. Rearrange the formula to find the molality: molality = 螖Tf / Kf Now substitute the known values to find the molality: molality = 5.00掳C / 1.86 掳C/molal molality 鈮� 2.69 molal
03

Determine the moles of ethylene glycol needed to achieve the desired molality

We know that molality is the number of moles of solute per kilogram of solvent. In this case, we have 1.00 kg of solvent (water) and want to achieve a molality of 2.69 molal. moles of ethylene glycol = molality 脳 mass of solvent moles of ethylene glycol = 2.69 molal 脳 1.00 kg moles of ethylene glycol 鈮� 2.69 mol
04

Convert the moles of ethylene glycol to grams

To find the mass of ethylene glycol needed, we need to multiply the moles of ethylene glycol by its molar mass. The molar mass of ethylene glycol (C2H6O2) is 62.07 g/mol. mass of ethylene glycol = moles of ethylene glycol 脳 molar mass of ethylene glycol mass of ethylene glycol = 2.69 mol 脳 62.07 g/mol mass of ethylene glycol 鈮� 166.96 g Therefore, approximately 166.96 grams of ethylene glycol must be added to 1.00 kg of water to produce a solution that freezes at -5.00掳C.

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Most popular questions from this chapter

By referring to Figure 13.15, determine whether the addition of \(40.0 \mathrm{~g}\) of each of the following ionic solids to \(100 \mathrm{~g}\) of water at \(40^{\circ} \mathrm{C}\) will lead to a saturated solution: (a) \(\mathrm{NaNO}_{3}\), (b) \(\mathrm{KCl}_{\text {, }}\) (c) \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\), (d) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\)

A solution is made containing \(20.8 \mathrm{~g}\) of phenol \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\right)\) in \(425 \mathrm{~g}\) of ethanol \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\right)\). Calculate (a) the mole fraction of phenol, (b) the mass percent of phenol, (c) the molality of phenol.

What is the freezing point of an aqueous solution that boils at \(105.0^{\circ} \mathrm{C}\) ?

The solubility of \(\mathrm{MnSO}_{4} \cdot \mathrm{H}_{2} \mathrm{O}\) in water at \(20^{\circ} \mathrm{C}\) is \(70 \mathrm{~g}\) per \(100 \mathrm{~mL}\) of water. (a) Is a \(1.22 \mathrm{M}\) solution of \(\mathrm{MnSO}_{4} \cdot \mathrm{H}_{2} \mathrm{O}\) in water at \(20^{\circ} \mathrm{C}\) saturated, supersaturated, or unsaturated? (b) Given a solution of \(\mathrm{MnSO}_{4} \cdot \mathrm{H}_{2} \mathrm{O}\) of unknown concentration, what experiment could you perform to determine whether the new solution is saturated, supersaturated, or unsaturated?

Are gases always miscible with each other? Explain. [Section 13.1]
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