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Chapter 16: Problem 36

To \(500 \mathrm{~cm}^{3}\) of water, \(3.0 \times 10^{-3} \mathrm{~kg}\) of acetic acid is added. If \(23 \%\) of acetic acid is dissociated, what will be the depression in freezing point? \(K_{f}\) and density of water are \(1.86 \mathrm{~K} \mathrm{~kg}^{1} \mathrm{~mol}^{1}\) and \(0.997 \mathrm{~g} \mathrm{~cm}^{3}\), respectively.

Short Answer

Expert verified
The depression in freezing point is 0.23 K.

Step by step solution

01

Convert mass of water to moles

First, let's determine the mass of water. Volume is given as 500 cm鲁 and the density is 0.997 g/cm鲁, so the mass is \(500 \times 0.997 = 498.5\) g or 0.4985 kg. Now, convert this mass to moles using the molar mass of water (18.015 g/mol): \[\text{moles of water} = \frac{498.5}{18.015} = 27.67\, \text{mol}\]
02

Convert acetic acid mass to moles

Given that 3.0 \(\times 10^{-3}\) kg (or 3.0 g) of acetic acid is added, we need to find the moles of acetic acid using its molar mass (60.05 g/mol):\[\text{moles of acetic acid} = \frac{3.0}{60.05} = 0.05\, \text{mol}\]
03

Calculate moles of dissociated acetic acid

Since 23% of acetic acid is dissociated, calculate the moles dissociated: \[\text{moles of dissociated acetic acid} = 0.23 \times 0.05 = 0.0115\, \text{mol}\]
04

Determine the van 't Hoff factor (i)

When acetic acid dissociates (CH鈧僀OOH 鈫� CH鈧僀OO鈦� + H鈦�), the van 't Hoff factor \(i\) accounts for dissociation:\[i = 1 + \alpha = 1 + 0.23 = 1.23\]
05

Calculate molality of the solution

Molality (m) is the number of moles of solute per kilogram of solvent. Here, the solvent is water:\[\text{molality} = \frac{\text{moles of acetic acid}}{\text{mass of water in kg}} = \frac{0.05}{0.4985} = 0.1003\, \text{mol/kg}\]
06

Calculate the depression in freezing point

The freezing point depression can be calculated using the formula \(\Delta T_f = i \, K_f \, m\):\[\Delta T_f = 1.23 \times 1.86 \, \text{K kg/mol} \times 0.1003 \, \text{mol/kg} = 0.2298 \, \text{K}\]
07

Final answer

The depression in freezing point is approximately 0.2298 K, indicating the freezing point will be lowered by this amount.

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Key Concepts

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

Van 't Hoff Factor
The Van 't Hoff factor, denoted as \( i \), is a key concept in colligative properties, which are properties of solutions that depend on the number of dissolved particles. The Van 't Hoff factor accounts for the dissociation of solutes in a solution.
The formula for calculating the Van 鈥榯 Hoff factor is \( i = 1 + \alpha \), where \( \alpha \) is the degree of dissociation. In the case of acetic acid (CH鈧僀OOH), it partially dissociates into ions \( \text{CH}_3\text{COO}^- \) and \( \text{H}^+ \).
This factor signifies how many effective particles result from the dissolution of one molecule of solute. For non-dissociating substances, \( i = 1 \). However, for substances like acetic acid that only partially dissociate, \( i \) can be a fractional value greater than 1. In our example, since \( 23\% \) of acetic acid dissociates, \( i = 1.23 \).
Understanding the Van 't Hoff factor helps in accurately calculating colligative properties like boiling point elevation and freezing point depression by considering actual particle count in the solution.
Molality
Molality is a measure of the concentration of a solute in a solution and is denoted by \( m \). It is defined as the number of moles of solute per kilogram of solvent. Unlike molarity, which is affected by changes in temperature and pressure, molality remains constant since it depends on mass, not volume.
To calculate molality, you use the formula: \[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]
In our context, acetic acid is our solute, while water is the solvent. With 0.05 moles of acetic acid dissolved in 0.4985 kg of water, the molality is 0.1003 mol/kg.
  • Molality is particularly useful in calculating colligative properties.
  • Because it remains unaffected by temperature, it is ideal for use in freezing point depression and boiling point elevation calculations.
By using molality, the errors that arise from using volume under varying conditions are eliminated, making it a more reliable measure for certain calculations.
Dissociation of Acetic Acid
Dissociation is the process where molecules split into smaller particles, such as ions. In solutions, this plays a big role in determining properties like conductivity and pH. Acetic acid, a weak acid represented by the formula CH鈧僀OOH, partially dissociates in water to form the ions CH鈧僀OO鈦� and H鈦�.
Since acetic acid is a weak acid, it does not fully dissociate in water. This partial dissociation is quantified by the degree of dissociation, denoted by \( \alpha \). In the given problem, \( 23\% \) dissociation means \( \alpha = 0.23 \).
The dissociation of acetic acid impacts colligative properties such as freezing point depression. When acetic acid dissociates, it increases the number of particles in the solution. This affects calculations using the Van 't Hoff factor, as previously discussed.
  • Only a portion of acetic acid molecules split into ions, influencing the overall behavior of the solution.
  • A higher degree of dissociation will result in a greater impact on the solution's freezing or boiling point.
Understanding dissociation helps in predicting and explaining the behavior of solutions in various chemical processes.

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Most popular questions from this chapter

How much amount of \(\mathrm{NaCl}\) should be added to \(600 \mathrm{~g}\) of water \((\mathrm{r}=\) \(1.00 \mathrm{~g} / \mathrm{mL}\) ) to decrease the freezing point of water to \(-0.2^{\circ} \mathrm{C}\) ? (The freezing point depression constant for water \(=2 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\) )

The vapour pressure of pure benzene at a certain temperature is 640 \(\mathrm{mm}\) Hg. A non-volatile non-electrolyte solid weighing \(2.175 \mathrm{~g}\) is added to \(39.0 \mathrm{~g}\) of benzene. The vapour pressure of the solution is 600 \(\mathrm{mm} \mathrm{Hg}\). What is the molecular weight of the solid substance?

\(1.22 \mathrm{~g}\) of benzoic acid is dissolved in \(100 \mathrm{~g}\) of acetone and \(100 \mathrm{~g}\) of benzene separately. Boiling point of the solution in acetone increases by \(0.17^{\circ} \mathrm{C}\), while that in the benzene increases by \(0.13^{\circ} \mathrm{C}\); \(K_{b}\) for acetone and benzene is \(1.7 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\) and \(2.6 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\) respectively. Find molecular weight of benzoic acid in two cases and justify your answer.

For an ideal solution of two components \(\mathrm{A}\) and \(\mathrm{B}\), which of the following is true? (a) \(\Delta \mathrm{H}_{\text {mixing }}<0\) (zero) (b) \(\Delta \mathrm{H}_{\text {mixing }}>0\) (zero) (c) \(\mathrm{A}-\mathrm{B}\) interaction is stronger than \(\mathrm{A}-\mathrm{A}\) and \(\mathrm{B}-\mathrm{B}\) interactions (d) \(\mathrm{A}-\mathrm{A}, \mathrm{B}-\mathrm{B}\) and \(\mathrm{A}-\mathrm{B}\) interactions are identical.

On dissolving \(0.5 \mathrm{~g}\) of a non-volatile non-ionic solute to \(39 \mathrm{~g}\) of benzene, its vapour pressure decreases from \(650 \mathrm{~mm} \mathrm{Hg}\) to \(640 \mathrm{~mm}\) \(\mathrm{Hg}\). The depression of freezing point of Benzene (in \(\mathrm{K}\) ) upon addition of the solute is (Given data : Molar mass and the molal freezing point depression constant of benzene are \(78 \quad \mathrm{~g} \quad \mathrm{~mol}^{-1} \quad\) and \(5.12 \quad \mathrm{~K} \quad \mathrm{~kg} \quad \mathrm{~mol}^{-1}\), respectively)
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