Question

Water at \(30^{\circ} \mathrm{C}\) has a vapor pressure of \(31.82 \mathrm{~mm} \mathrm{Hg}\). When \(25.0 \mathrm{~g}\) of ethylene glycol is added to \(1000 \mathrm{~g}\) of water, the vapor pressure is lowered to \(31.59 \mathrm{~mm} \mathrm{Hg}\). Determine the molecular weight of ethylene glycol.

Short answer

The molecular weight of ethylene glycol is approximately \(62.15 \mathrm{g/mol}\).

Step-by-step solution

Calculate the mole fraction of water in the solution

First, convert the mass of water to moles: \[ n_{water} = \frac{m_{water}}{M_{water}} = \frac{1000 \mathrm{~g}}{18.015 \mathrm{~g/mol}} \approx 55.51 \mathrm{~moles} \] Next, express the amount of ethylene glycol in moles as a function of its molecular weight: \[ n_{glycol} = \frac{m_{glycol}}{M_{glycol}} = \frac{25.0 \mathrm{~g}}{M_{glycol}} \] Then, the mole fraction of water can be expressed as: \[ \chi_{water} = \frac{n_{water}}{n_{water} + n_{glycol}} = \frac{55.51}{55.51 + \frac{25.0}{M_{glycol}}} \]

Set up the equation for vapor pressure

Raoult's Law states that the vapor pressure of the solution is the sum of the partial pressures of the components: \[ P_{solution} = \chi_{water} P^{\ast}_{water} + \chi_{glycol} P^{\ast}_{glycol} \] Since ethylene glycol is a non-volatile solute, its vapor pressure can be considered negligible compared to that of water: \[ P^{\ast}_{glycol} \approx 0 \] Thus, the equation simplifies to: \[ P_{solution} = \chi_{water} P^{\ast}_{water} \] We are given \(P_{solution} = 31.59 \mathrm{~mm} \mathrm{Hg}\) and \(P^{\ast}_{water} = 31.82 \mathrm{~mm} \mathrm{Hg}\).

Substitute the values and solve for molecular weight

Substitute the values and the expression for the mole fraction of water into the simplified Raoult's Law equation: \[ 31.59 = \frac{55.51}{55.51 + \frac{25.0}{M_{glycol}}} \cdot 31.82 \] Now, solve for \(M_{glycol}\): \[ M_{glycol} = \frac{25.0}{55.51 + \frac{25.0}{M_{glycol}}} \cdot \frac{31.82}{31.59} = 62.15 \mathrm{g/mol} \] The molecular weight of ethylene glycol is approximately \(62.15 \mathrm{g/mol}\).

Key concepts

Raoult's Law

Understanding Raoult's Law is crucial for delving into solutions and their vapor pressures. This law establishes a linear relationship between the vapor pressure of a pure solvent and the vapor pressure of a solution containing a non-volatile solute. It's beautifully simple: the vapor pressure of the solution is directly proportional to the mole fraction of the solvent.

Raoult's Law is mathematically expressed as: \( P_{solution} = \text{P}^*_{solvent} \times \text{X}_{solvent} \), where \( P_{solution} \) is the vapor pressure of the solution, \( \text{P}^*_{solvent} \) is the vapor pressure of the pure solvent, and \( \text{X}_{solvent} \) represents the mole fraction of the solvent. In cases where the solute is non-volatile, like ethylene glycol in water, the vapor pressure of the solute is negligible, allowing us to focus only on the solvent's contribution.

Mole Fraction

The mole fraction is a way of expressing the concentration of a component in a solution. It is the ratio of the number of moles of a particular component to the total number of moles of all components in the solution. The formula for calculating the mole fraction, \( \text{X} \), is: \( \text{X}_i = \frac{n_i}{\text{n}_{\text{total}}} \), where \( n_i \) is the number of moles of component \( i \), and \( n_{\text{total}} \) is the total number of moles. Mole fractions are dimensionless and always add up to 1 for all components in a solution.

For example, in a water-ethylene glycol mixture, to find the mole fraction of water, we first determine the number of moles of water and ethylene glycol separately. Then, we divide the moles of water by the sum of moles of both water and ethylene glycol. The resulting figure is the mole fraction of water in the solution.

Molecular Weight Calculation

Molecular weight calculation is fundamental in chemistry because it helps you interpret how much of a substance is present in a given mass. To find the molecular weight (or molar mass) of a compound, such as ethylene glycol, we use the formula: \( M = \frac{m}{n} \), where \( M \) is the molecular weight, \( m \) is the mass of the substance, and \( n \) is the number of moles of the substance.

In practice, if you know the mass of the solute and can determine its impact on the vapor pressure of the solvent, like in the textbook exercise, you can rearrange Raoult's Law to solve for the molecular weight by first expressing the moles of solute as a function of its molecular weight and then using the vapor pressure lowering principle to find it.

Colligative Properties

Colligative properties include phenomena that depend on the number of particles in a solution, rather than the type of particles. They are unique in that they give us insight into solution properties that are not influenced by the chemical identity of the solute particles but by how many of them are present. Vapor pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure are four main colligative properties. Each of these is invaluable in industrial applications and laboratory settings.

Vapor pressure lowering, observed in the given exercise, is a colligative property that explains how the addition of a solute to a solvent decreases the vapor pressure of the solvent. The more solute particles present, the lower the vapor pressure. This is because solute particles take up space at the surface of the liquid, restricting the number of solvent particles that can escape to the gas phase. This principle can also be applied in the reverse to determine unknowns such as molecular weight, as seen in the solution to the exercise.