Problem 98 - Chemistry and Chemical Reactivity

Question

Which substance would have the greater influence on the vapor pressure of water when added to \(1000 .\) g of the liquid: \(10.0 \mathrm{g}\) of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) or \(10.0 \mathrm{g}\) of ethylene glycol \(\left(\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}\right) ?\)

Step-by-Step Solution

Answer

Ethylene glycol has a greater influence.

Step 1: Understand the Concept

The vapor pressure of a solvent decreases when a non-volatile solute is added to it. This is explained by Raoult's law which states that the relative lowering of vapor pressure is proportional to the mole fraction of the solute.

Step 2: Calculate Moles of Sucrose

Find the molar mass of sucrose (C₁₂H₂₂O₁₁). The molar mass is calculated as: \(12(12) + 22(1) + 11(16) = 342 \ \text{g/mol}\). Then calculate the moles of sucrose: \(\text{moles} = \frac{10.0}{342} = 0.0292 \ \text{mol}\).

Step 3: Calculate Moles of Ethylene Glycol

Find the molar mass of ethylene glycol (C₂H₆O₂). The molar mass is calculated as: \(2(12) + 6(1) + 2(16) = 62 \ \text{g/mol}\). Then calculate the moles of ethylene glycol: \(\text{moles} = \frac{10.0}{62} = 0.1613 \ \text{mol}\).

Step 4: Compare Amount of Substance

The influence on vapor pressure depends on the number of moles of solute, not their mass. More moles cause a greater decrease in vapor pressure. Ethylene glycol has more moles (0.1613 mol) compared to sucrose (0.0292 mol).

Step 5: Conclude the Greater Influence

Since ethylene glycol has more moles, it will have a greater influence on lowering the vapor pressure of water compared to sucrose when 10.0 g of each is dissolved in 1000 g of water.

Key Concepts

Vapor Pressure

Vapor pressure is a crucial concept in understanding how substances behave in solutions. It refers to the pressure exerted by the vapor present above a liquid surface. This pressure comes from molecules that have enough kinetic energy to escape the liquid phase and enter the gas phase. The thing about vapor pressure is that it is dependent on the temperature. Warmer temperatures mean more molecules can escape, thus increasing the vapor pressure.

When you add a solute to a solvent, if the solute is non-volatile, it doesn't add to the vapor pressure itself. Instead, it reduces the number of solvent molecules at the liquid's surface that can escape into the vapor phase, thus lowering the vapor pressure. This is an essential part of Raoult's Law, which describes the relationship between solute and vapor pressure changes.

Non-Volatile Solute

A non-volatile solute is one that does not vaporize readily. Unlike alcohol or other volatile compounds, these solutes stay in the liquid phase. When added to a solvent, they cause a reduction in the solvent's vapor pressure.

By reducing the vapor pressure, non-volatile solutes make it less likely for the solvent molecules to escape into the gaseous phase. This phenomenon is the foundation of many practical applications, including boiling point elevation and freezing point depression -- both are colligative properties. Here, we need to consider how the number of particles of the non-volatile solute relates to the effect on vapor pressure, which is crucial for solving problems such as the one given.

Mole Fraction

The mole fraction is a way to express the concentration of a component in a mixture. It is the ratio of the number of moles of a specific component to the total number of moles in the solution. The formula is simple:

• Mole Fraction ( \(x_i\)) = \(\frac{n_i}{n_{\text{total}}}\), where \(n_i\) is the moles of the component, and \(n_{\text{total}}\) is the total moles in the mixture.

In terms of vapor pressure, Raoult's Law states that the partial vapor pressure of each component in a solution is equal to the vapor pressure of the pure component multiplied by its mole fraction. So, a higher mole fraction of a solute would mean a greater effect on reducing the vapor pressure, making understanding its calculation critical in exercises involving Raoult's Law.

Molar Mass Calculation

Molar mass calculation is a fundamental skill required in most chemistry problems. It involves finding the sum of the atomic masses of all atoms in a molecule. For instance, when calculating the molar mass of sucrose (\( \text{C}_{12} \text{H}_{22} \text{O}_{11} \)), you add:

• 12 carbons: \(12 \times 12 \text{ g/mol} = 144 \text{ g/mol}\)
• 22 hydrogens: \(22 \times 1 \text{ g/mol} = 22 \text{ g/mol}\)
• 11 oxygens: \(11 \times 16 \text{ g/mol} = 176 \text{ g/mol}\)

Putting them together, sucrose's molar mass is \(342 \text{ g/mol}\). Ethylene glycol, with its simpler structure, has different molar masses: 2 carbons, 6 hydrogens, and 2 oxygens leading to a molar mass of \(62 \text{ g/mol}\).

In the original exercise, calculating the number of moles for each substance using their molar masses allows us to determine which has more moles and, ultimately, which has a greater effect on lowering the vapor pressure of water.