Question

The vapor pressure of a solution containing 53.6 g glycerin \(\left(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{3}\right)\) in \(133.7 \mathrm{g}\) ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) is 113 torr at \(40^{\circ} \mathrm{C}\) Calculate the vapor pressure of pure ethanol at \(40^{\circ} \mathrm{C}\) assuming that glycerin is a nonvolatile, nonelectrolyte solute in ethanol.

Short answer

The vapor pressure of pure ethanol at 40°C is 135.7 torr.

Step-by-step solution

Determine the mole fraction of ethanol in the solution

First, calculate the moles of each component in the solution: - Molar mass of glycerin (C3H8O3): 3(12.01) + 8(1.01) + 3(16.00) = 92.09 g/mol - Moles of glycerin: 53.6 g / 92.09 g/mol = 0.582 moles - Molar mass of ethanol (C2H5OH): 2(12.01) + 6(1.01) + 1(16.00) = 46.07 g/mol - Moles of ethanol: 133.7 g / 46.07 g/mol = 2.90 moles Next, determine the mole fraction (x) of ethanol in the solution: - x_ethanol = moles of ethanol / (moles of ethanol + moles of glycerin) = 2.90 / (2.90 + 0.582) = 0.833

Use Raoult's Law to calculate the vapor pressure of pure ethanol

Raoult's Law states that the vapor pressure of a volatile component (ethanol) in a mixture is equal to the product of the mole fraction of that component and its vapor pressure in the pure state: - P_solution = x_ethanol * P_ethanol_pure We know the vapor pressure of the solution (P_solution) is 113 torr and the mole fraction of ethanol (x_ethanol) is 0.833. Solving for P_ethanol_pure: - P_ethanol_pure = P_solution / x_ethanol - P_ethanol_pure = 113 torr / 0.833 = 135.7 torr

Report the vapor pressure of pure ethanol at 40°C

The vapor pressure of pure ethanol at 40°C is 135.7 torr.

Key concepts

Understanding Mole Fraction

In chemistry, mole fraction is a way to express the concentration of a component in a mixture. It's the ratio of the number of moles of a particular substance to the total number of moles of all substances present. Simply put, it's a way to describe how much of a particular substance exists compared to the whole mix.

For example, if you have a solution of glycerin and ethanol, you calculate the moles of each component and then divide the moles of one component (in this case, ethanol) by the total moles. This gives you the mole fraction of ethanol in the solution. It is a dimensionless number and always less than or equal to one because it represents a part of a whole.

Raoult's Law in Vapor Pressure Calculation

Raoult's Law is a fundamental principle in physical chemistry, particularly essential when studying solutions. It defines the relationship between the vapor pressure of a solvent and the presence of a nonvolatile solute. According to this law, the vapor pressure of a solvent above a solution (containing a nonvolatile solute) is directly proportional to the mole fraction of the solvent in the solution.

The formula for Raoult's Law is expressed as:

Principle of Raoult's Law

\( P_{solution} = x_{solvent} \times P^{\circ}_{solvent} \)

Here, \( P_{solution} \) is the vapor pressure of the solution, \( x_{solvent} \) is the mole fraction of the solvent, and \( P^{\circ}_{solvent} \) is the vapor pressure of the pure solvent. By applying this law, we can determine the influence of solutes on the vapor pressure of solutions, facilitating the calculation of the vapor pressure of pure substances in mixtures.

Role of Nonvolatile Solute in Solutions

A nonvolatile solute is a substance that does not readily evaporate into a gas under existing conditions. In a solution, the presence of a nonvolatile solute decreases the vapor pressure relative to the pure solvent. This occurs because the solute molecules occupy space at the surface of the liquid, thus reducing the number of solvent molecules that can escape into the vapor phase.

Nonvolatile solutes, therefore, play a crucial role in applications like antifreeze in car radiators or the preservation of food where the solution’s boiling or freezing points are manipulated. Because these solutes do not add any vapor pressure of their own, they allow for straightforward applications of Raoult's Law, as illustrated in the solution to our textbook problem regarding glycerin and ethanol.