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Chapter 5: Problem 51

The vapours pressure of water at \(23^{\circ} \mathrm{C}\) is \(19.8 \mathrm{~mm}\). of \(\mathrm{Hg} .0 .1\) mole of glucose is dissolved in \(178.2 \mathrm{~g}\) of water. What is the vapour pressure (in \(\mathrm{mm}\) of \(\mathrm{Hg}\) ) of the resultant solution? (a) \(19.0\) (b) \(19.602\) (c) \(19.402\) (d) \(19.202\)

Short Answer

Expert verified
The vapour pressure of the solution is 19.602 mmHg.

Step by step solution

01

Understanding the Formula for Vapour Pressure Lowering

We will use Raoult's Law, which states that the vapour pressure of the solution is the vapour pressure of the pure solvent times the mole fraction of the solvent in the solution. The formula can be expressed as \( P_{solution} = P_{solvent} \times \chi_{solvent} \), where \( P_{solution} \) is the vapour pressure of the solution, \( P_{solvent} \) is the vapour pressure of the pure solvent, and \( \chi_{solvent} \) is the mole fraction of the solvent.
02

Calculate the Moles of Water

Given that the mass of water is \(178.2 \text{ g}\) and the molar mass of water is \(18.02 \text{ g/mol}\), we can calculate the moles of water. \[ \text{Moles of water} = \frac{178.2}{18.02} \approx 9.89 \text{ moles} \]
03

Determine the Mole Fraction of the Solvent

To find the mole fraction of water, use the formula \( \chi_{solvent} = \frac{n_{water}}{n_{water} + n_{glucose}} \), where \( n_{water} \) is the moles of water and \( n_{glucose} \) is the moles of glucose, which is given as 0.1 mole. \[ \chi_{solvent} = \frac{9.89}{9.89 + 0.1} \approx 0.99 \]
04

Apply Raoult's Law to Find Vapour Pressure

Using the calculated mole fraction of water in the solution and the initial vapour pressure of pure water \( P_{solvent} = 19.8 \text{ mmHg} \), apply the formula: \[ P_{solution} = 19.8 \times 0.99 \approx 19.602 \text{ mmHg} \]
05

Conclusion

The vapour pressure of the resultant solution is determined to be approximately \(19.602 \text{ mmHg}\). Thus, the correct answer is option (b).

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Key Concepts

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

Vapour Pressure
Vapour pressure is a key concept in solution chemistry. It refers to the pressure exerted by the vapour of a liquid in equilibrium with its liquid phase at a given temperature. This equilibrium signifies that the rate of evaporation of the liquid equals the rate of condensation of the vapour.
In simpler terms, vapour pressure is the tendency of molecules to escape into the gas phase. For pure liquids, the vapour pressure is a characteristic value at each particular temperature. However, when a non-volatile solute is dissolved in a liquid, the vapour pressure of the resulting solution is typically lower than that of the pure solvent.
  • The decrease in vapour pressure is due to solute molecules occupying surface space, preventing some solvent molecules from escaping into the vapour phase.
  • Raoult's Law helps us quantify this change in vapour pressure for ideal solutions.
Mole Fraction
The mole fraction is a measure that helps us understand the composition of a mixture or a solution. It is defined as the ratio of the number of moles of a particular component to the total number of moles in the mixture.
In a solution, mole fraction provides insight into the concentration of the components without involving units like molarity, which depend on volume. We calculate the mole fraction as follows:
  • Formula: \[ \chi_{i} = \frac{n_{i}}{n_{total}} \]where \( \chi_{i} \) represents the mole fraction of component \( i \), \( n_{i} \) is the number of moles of component \( i \), and \( n_{total} \) is the total number of moles of all components in the solution.
  • In our problem, the mole fraction is calculated for water (the solvent), considering both water and glucose (the solute).
The mole fraction closely ties into Raoult's Law, as it is used to determine how the presence of solute affects the solvent's vapour pressure.
Solution Chemistry
Solution chemistry is the study of the properties and behavior of solutions, focusing significantly on how solutes and solvents interact. A solution consists of two major parts: the solute (the substance being dissolved) and the solvent (the substance doing the dissolving).
In this scenario, we are dealing with a solution made by dissolving glucose in water. Understanding solution chemistry involves several important concepts:
  • Solubility: The ability of a solute to dissolve in a solvent at a specific temperature and pressure.
  • Concentration: Indicates how much solute is present in a certain quantity of solvent or solution, expressible in units like molarity, molality, or as in this problem, mole fraction.
  • Colligative Properties: Properties that depend on the number of solute particles in a solution, not the identity of solute particles. Vapour pressure lowering is one such property.
These concepts help us understand why the presence of glucose in the water affects the vapour pressure, illustrating how solute-solvent interactions lead to real-world phenomena like boiling point elevation and freezing point depression.

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Most popular questions from this chapter

A binary liquid solution is prepared by mixing n-heptane and ethanol. Which on of the following statement is correct regarding the behavior of the solution? (a) The solution in non-ideal, showing +ve deviation from Raoult's Law. (b) The solution in non-ideal, showing -ve deviation from Raoult's Law. (c) n-heptane shows tre deviation while ethanol shows -ve deviation from Raoult's Law. (d) The solution formed is an ideal solution.

What would be the \(\mathrm{pH}\) of a \(0.1\) molal aqueous solution of a monoprotic acid 'HA', that freezes at \(-0.2046^{\circ} \mathrm{C}\) ? \(\left[\mathrm{K}_{\mathrm{f}}\left(\mathrm{H}_{2} \mathrm{O}\right)=1.86^{\circ} \mathrm{mol}^{-1} \mathrm{~kg} ;\right.\) assuming molality \(=\) molarity]

For an aqueous solution, freezing point is \(-0.186^{\circ} \mathrm{C}\). Elevation of the boiling point of the same solution is $$ \left(\mathrm{K}_{\ell}=1.86^{\circ} \mathrm{mol}^{-1} \mathrm{~kg} \text { and } \mathrm{K}_{\mathrm{b}}=0.512^{\circ} \mathrm{mol}^{-1} \mathrm{~kg}\right) $$ (a) \(0.186^{\circ}\) (b) \(0.0512^{\circ}\) (c) \(1.86^{\circ}\) (d) \(5.12^{\circ}\)

In a \(0.2\) molal aqueous solution of a weak acid HX, the degree of ionization is \(0.3 .\) Taking \(K_{f}\) for water as \(1.85 \mathrm{k} \mathrm{kg}\) melt, the freezing point of the solution will be nearest to (a) \(-0.480^{\circ} \mathrm{C}\) (b) \(-0.360^{\circ} \mathrm{C}\) (c) \(-0.260^{\circ} \mathrm{C}\) (d) \(+0.480^{\circ} \mathrm{C}\)

The amount of dissolved oxygen in 1 litre water in equilibrium with air at 1 atm pressure at \(25^{\circ} \mathrm{C}\) will be (assume that air contains 20 mole \% oxygen, Henry's constant \((\mathrm{kH})\) for oxygen is \(3.04 \times 10^{7} \mathrm{~mm} \mathrm{Hg}\) and density of \(\mathrm{H}_{2} \mathrm{O}\) at \(25^{\circ} \mathrm{C}\) is \(1 \mathrm{~g} / \mathrm{cc}\) ) (a) the mole fraction of \(\mathrm{O}_{2}\) in solution is \(5 \times 10^{-6}\) (b) the mole fraction of \(\mathrm{O}_{2}^{2}\) in solution is \(5 \times 10^{-7}\) (c) the molarity of \(\mathrm{O}_{2}\) in solution is \(2.77 \times 10^{-5} \mathrm{M}\) (d) the molarity of \(\mathrm{O}_{2}\) in solution is \(2.77 \times 10^{4} \mathrm{M}\)
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