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Chapter 14: Problem 80

Which would be more effective as an antifreeze in an automobile radiator? A solution containing (a) \(10 \mathrm{~kg}\) of methyl alcohol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) or \(10 \mathrm{~kg}\) of ethyl alcohol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right) ?\) (b) \(10 \mathrm{~m}\) solution of methyl alcohol or \(10 \mathrm{~m}\) solution of ethyl alcohol?

Short Answer

Expert verified
(a) Methyl alcohol is more effective. (b) Both are equally effective.

Step by step solution

01

Determine molar masses

Calculate the molar masses of methyl alcohol \(\text{CH}_3\text{OH}\) and ethyl alcohol \(\text{C}_2\text{H}_5\text{OH}\). The molar mass of methyl alcohol is \(12 + (3 \times 1) + 16 + 1 = 32 \text{g/mol}\), and the molar mass of ethyl alcohol is \(2 \times 12 + (6 \times 1) + 16 = 46\text{g/mol}\).
02

Convert mass to moles for part (a)

For part (a), convert 10 kg of methyl alcohol to moles: \(\frac{10,000 \text{g}}{32 \text{g/mol}} = 312.5 \text{mol}\). Similarly, convert 10 kg of ethyl alcohol to moles: \(\frac{10,000 \text{g}}{46 \text{g/mol}} = 217.4 \text{mol}\).
03

Compare moles for part (a)

For part (a), compare the number of moles: 312.5 moles of methyl alcohol and 217.4 moles of ethyl alcohol. More moles mean a higher concentration, leading to a lower freezing point.
04

Analyze molality for part (b)

For part (b), both solutions have the same molality (10 m). The effect on the freezing point depends on the concentration of solute particles, which is the same for both, making them equally effective.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molar Mass Calculation
To determine which alcohol is more effective as an antifreeze, we start by calculating the molar masses of the substances involved. Knowing the molar mass helps us understand how many moles are present in a given mass of the substance.

For methyl alcohol \(\text{CH}_3\text{OH}\), we calculate:
  • Carbon (C): 1 atom \( \times 12 \text{g/mol} = 12 \text{g/mol} \)
  • Hydrogen (H): 4 atoms \( \times 1 \text{g/mol} = 4 \text{g/mol} \)
  • Oxygen (O): 1 atom \( \times 16 \text{g/mol} = 16 \text{g/mol} \)
Sum these together to get a molar mass of 32 \text{g/mol}.

For ethyl alcohol \(\text{C}_2\text{H}_5\text{OH}\), we proceed similarly:
  • Carbon (C): 2 atoms \( \times 12 \text{g/mol} = 24 \text{g/mol} \)
  • Hydrogen (H): 6 atoms \( \times 1 \text{g/mol} = 6 \text{g/mol} \)
  • Oxygen (O): 1 atom \( \times 16 \text{g/mol} = 16 \text{g/mol} \)
Summing these gives a molar mass of 46 \text{g/mol}.

These calculations set the stage for converting mass to moles.
Molecular Conversion
Next, we need to convert the given mass of each alcohol into moles. This conversion is crucial to compare how many molecules of each type are present.

For part (a) of the problem, we have 10 kg of each alcohol to consider.
  • For methyl alcohol, convert 10,000g to moles: \( \frac{10,000 \text{g}}{32 \text{g/mol}} = 312.5 \text{mol} \)
  • For ethyl alcohol, the calculation is: \( \frac{10,000 \text{g}}{46 \text{g/mol}} = 217.4 \text{mol} \)
We see that 10 kg of methyl alcohol contains 312.5 moles, while 10 kg of ethyl alcohol contains 217.4 moles.

More moles of solute usually indicate that the solution will have a greater effect on the freezing point depression.
Freezing Point Depression
Freezing point depression refers to the phenomenon where the addition of a solute to a solvent decreases the freezing point of the solvent. This principle is used to enhance the effectiveness of antifreeze.

According to the formula: \(\text{Δ}T_f = i \times K_f \times \text{molality} \), the decrease in freezing point \( \text{Δ}T_f \) depends on:
  • \text{van 't Hoff factor} (i): Number of particles the solute splits into \
  • \text{Freezing point depression constant} (K_f): A property of the solvent \
  • \text{Molality}: The concentration of solute particles \
For methyl alcohol and ethyl alcohol, both \(i\) is 1 (since they don’t dissociate in water) and \(K_f\) is constant for water. Thus, the one with more moles (methyl alcohol in part (a)) results in a greater freezing point depression and is more effective as an antifreeze.
Molality
Molality is a measure used to describe the concentration of a solution. It's defined as the number of moles of solute per kilogram of solvent. In this scenario, it's critical to understanding the impact of each alcohol on freezing point.

For part (b), we must consider solutions with the same molality.
  • A 10 m solution of methyl alcohol
  • A 10 m solution of ethyl alcohol
Because both solutions have the same molality, they contribute equally to freezing point depression. No matter what type of alcohol is used, their molality being equal means that they will be equally effective as antifreeze in this concentration, provided the solute particles don't dissociate or interact in a complex way.

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Most popular questions from this chapter

A sugar syrup solution contains \(15.0 \%\) sugar, \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\), by mass and has a density of \(1.06 \mathrm{~g} / \mathrm{mL}\). (a) How many grams of sugar are in \(1.0 \mathrm{~L}\) of this syrup? (b) What is the molarity of this solution? (c) What is the molality of this solution?

Automobile battery acid is \(38 \% \mathrm{H}_{2} \mathrm{SO}_{4}\) and has a density of \(1.29 \mathrm{~g} / \mathrm{mL}\). Calculate the molality and the molarity of this solution.

Determine the volume percent of a solution made by dissolving: (a) \(37.5 \mathrm{~mL}\) of butanol in enough ethanol to make \(275 \mathrm{~mL}\) of solution (b) \(4.0 \mathrm{~mL}\) of methanol in enough water to make \(25.0 \mathrm{~mL}\) of solution (c) \(45.0 \mathrm{~mL}\) of isoamyl alcohol in enough acetone to make \(750 . \mathrm{mL}\) of solution

For each pairing below, predict which will dissolve faster and explain why. (a) \(0.424 \mathrm{~g}\) of the amino acid phenylalanine in \(50.0 \mathrm{~g}\) of isopropyl alcohol at \(25^{\circ} \mathrm{C}\) or in \(50.0 \mathrm{~g}\) of isopropyl alcohol at \(75^{\circ} \mathrm{C}\) (b) a \(1.42\)-gcrystal of sodium acetate in \(300 \mathrm{~g}\) of \(37^{\circ} \mathrm{C}\) water or \(1.42 \mathrm{~g}\) of powdered sodium acetate in \(300 \mathrm{~g}\) of \(37^{\circ} \mathrm{C}\) water (c) a \(500-\mathrm{g}\) carton of table salt \((\mathrm{NaCl})\) in \(5.0 \mathrm{~L}\) of water at \(35^{\circ} \mathrm{C}\) or a 500 -g salt lick ( \(\mathrm{NaCl})\) for livestock in \(5.0 \mathrm{~L}\) of water at \(21^{\circ} \mathrm{C}\) (d) \(25 \mathrm{mg}\) of acetaminophen in \(100 \mathrm{~mL}\) of infant pain medication containing \(160 \mathrm{mg}\) of acetaminophen per \(15 \mathrm{~mL}\) or in \(100 \mathrm{~mL}\) of adult pain medication containing \(320 \mathrm{mg}\) of acetaminophen per \(15 \mathrm{~mL}\)

A bleaching solution requires \(5.23\) g of sodium hypochlorite. How many grams of a \(21.5 \%\) by mass solution of sodium hypochorite should be used?
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