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Chapter 16: Problem 37

A solution of a non-volatile solute in water freezes at \(-0.30^{\circ} \mathrm{C}\). The vapour pressure of pure water at \(298 \mathrm{~K}\) is \(23.51 \mathrm{~mm} \mathrm{Hg}\) and \(K_{f}\) for water is \(1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} .\) Calculate the vapour pressure of this solution at \(298 \mathrm{~K}\).

Short Answer

Expert verified
The vapour pressure of the solution at 298 K is approximately 23.44 mm Hg.

Step by step solution

01

Calculating the Molality (m)

We start by finding the molality of the solution using the formula for freezing point depression: \\[\Delta T_f = K_f \times m\] where \(\Delta T_f = 0.30^{\circ}C\) and \(K_f = 1.86 \, \text{K}\, \text{kg}\, \text{mol}^{-1}\). Thus: \\[m = \frac{\Delta T_f}{K_f} = \frac{0.30}{1.86}\approx 0.1613 \, \text{mol} \cdot \text{kg}^{-1}\]
02

Determining Mole Fraction of Solute

Assuming one kg of water, the moles of solute (from the molality) would be 0.1613 mol. The moles of water can be calculated as it has a molar mass of 18 g/mol: \\[\text{Moles of water} = \frac{1000}{18} \approx 55.56 \, \text{mol}\] \[\text{Total moles} = 55.56 + 0.1613 = 55.7213\] \[\text{Mole fraction of water} = \frac{55.56}{55.7213}\approx 0.9971\]
03

Calculating Vapour Pressure of Solution

Using Raoult's Law: \(P_{\text{solution}} = \text{Mole fraction of water} \times P^0_{\text{water}}\), where \(P^0_{\text{water}} = 23.51 \, \text{mm Hg}\). Thus: \\[P_{\text{solution}} = 0.9971 \times 23.51 \approx 23.44 \, \text{mm Hg}\]

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Key Concepts

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

Molality
Molality is a measure of the concentration of a solute in a solution. Unlike molarity, which is sensitive to the volume changes with temperature, molality is defined in terms of mass, making it a more stable measurement for various conditions. Here's a closer look at what molality actually entails:

  • Definition: Molality ( \( m \) ) is the number of moles of solute dissolved in one kilogram of solvent.
  • Formula: To calculate molality, the formula is: \[ m = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}} \]
  • Application: Molality is particularly useful for freezing point depression, boiling point elevation, and various other colligative properties. It gives an accurate measure as the solvent's mass does not change with temperature.
In the given exercise, the freezing point depression data were used to calculate molality, revealing the direct relationship between the solute concentration and the extent to which it lowers the freezing point. Such calculations are crucial in chemistry to prepare solutions with specific properties while considering changes due to environmental factors.
Raoult's Law
Raoult's Law is fundamental in understanding how the vapour pressure of a solvent is affected by adding a solute. It helps predict how a solution behaves differently from the pure solvent. Here's why Raoult's Law is significant in this context:

  • Basic Principle: The vapour pressure of a solution is directly proportional to the mole fraction of the solvent in the solution.
  • Mathematical Representation: Expressed through the formula: \[ P_{\text{solution}} = \text{Mole fraction of solvent} \times P^0_{\text{solvent}} \]
  • Importance: It shows that the presence of a non-volatile solute lowers the vapour pressure of the solution compared to the pure solvent.
For this exercise, Raoult's Law is used to determine the vapour pressure of the given solution of water and non-volatile solute. After finding the mole fraction of the solvent, which is water in this case, the law helps calculate the reduced vapour pressure, capturing the effect of dissolved solutes in practical terms.
Vapour Pressure
The concept of vapour pressure is crucial to understand when studying solutions, especially in relation to colligative properties like freezing point depression and boiling point elevation. Let's dive deeper into what vapour pressure means:

  • Definition: Vapour pressure is the pressure exerted by the vapour when a liquid is in dynamic equilibrium with its vapour. It is essentially a measure of a liquid's tendency to evaporate.
  • Factors Affecting Vapour Pressure:
    • Temperature: As temperature increases, vapour pressure increases because more molecules have adequate energy to escape into the gas phase.
    • Nature of the Liquid: Volatile liquids have higher vapour pressures than less volatile ones.
  • Impact of Solutes: When a non-volatile solute is added to a solvent like water, the vapour pressure decreases due to fewer solvent molecules at the surface able to escape into the vapour phase.
In the exercise, the goal was to calculate the vapour pressure of a solution at a specific temperature. This involves understanding how the addition of a solute modifies the physical properties of the solvent, directly impacting its vapour pressure and the overall behaviour of the solution.

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Most popular questions from this chapter

The mole fraction of a solvent in aqueous solution of a solute is \(0.8\). The molality (in \(\mathrm{mol} \mathrm{kg}^{-1}\) ) of the aqueous solution is : (a) \(13.88 \times 10^{-2}\) (b) \(13.88 \times 10^{-1}\) (c) \(13.88\) (d) \(13.88 \times 10^{-3}\)

Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogeneous solution. These are called colligative properties. Application of colligative properties are very useful in day-to-day life. One of its example is the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator of automobiles. A solution \(\mathrm{M}\) is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is \(0.9\) Given : Freezing point depression constant of water \(\left(K_{f}^{\text {water }}\right)\) $$ =1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} $$ Freezing point depression constant of ethanol ( \(\left.K_{f}{ }^{\text {ethanol }}\right)\) $$ =2.0 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} $$ Boiling point elevation constant of water \(\left(K_{b}^{\text {water }}\right)\) \(=0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\) Boiling point elevation constant of ethanol \(\left(K_{b}^{\text {ethanol }}\right)=1.2 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\) Standard freezing point of water \(=273 \mathrm{~K}\) Standard freezing point of ethanol \(=155.7 \mathrm{~K}\) Standard boiling point of water \(=373 \mathrm{~K}\) Standard boiling point of ethanol \(=351.5 \mathrm{~K}\) Vapour pressure of pure water \(=32.8 \mathrm{~mm} \mathrm{Hg}\) Vapour pressure of pure ethanol \(=40 \mathrm{~mm} \mathrm{Hg}\) Molecular weight of water \(=18 \mathrm{~g} \mathrm{~mol}^{-1}\) Molecular weight of ethanol \(=46 \mathrm{~g} \mathrm{~mol}^{-1}\) In answering the following questions, consider the solution to be ideal dilute solutions and solutes to be non-volatile and non-dissociative. The freezing point of the solution \(\mathrm{M}\) is (a) \(268.7 \mathrm{~K}\) (b) \(268.5 \mathrm{~K}\) (c) \(234.2 \mathrm{~K}\) (d) \(150.9 \mathrm{~K}\)

A compound \(\mathrm{H}_{2} X\) with molar weight of \(80 \mathrm{~g}\) is dissolved in a solvent having density of \(0.4 \mathrm{~g} \mathrm{~mL}^{-1}\). Assuming no change in volume upon dissolution, the molality of a 3.2 molar solution is _____ .

During depression of freezing point in a solution, the following are in equililbrium (a) liquid solvent, solid solvent (b) liquid solvent, solid solute (c) liquid solute, solid solute (d) liquid solute, solid solvent

At \(300 \mathrm{~K}\), the vapour pressure of a solution containing 1 mole of \(n\) hexane and 3 moles of \(n\)-heptane is \(550 \mathrm{~mm}\) of \(\mathrm{Hg}\). At the same temperature, if one more mole of \(n\)-heptane is added to this solution, the vapour pressure of the solution increases by \(10 \mathrm{~mm}\) of \(\mathrm{Hg}\). What is the vapour pressure in \(\mathrm{mm} \mathrm{Hg}\) of \(n\)-heptane in its pure state _____ ?
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