Question

A nonvolatile organic compound \(Z\) was used to make up two solutions. Solution A contains \(5.00 \mathrm{~g}\) of \(Z\) dissolved in \(100 \mathrm{~g}\) of water, and solution \(\mathrm{B}\) contains \(2.31 \mathrm{~g}\) of \(Z\) dissolved in \(100 \mathrm{~g}\) of benzene. Solution A has a vapor pressure of \(754.5 \mathrm{mmHg}\) at the normal boiling point of water, and solution \(\mathrm{B}\) has the same vapor pressure at the normal boiling point of benzene. Calculate the molar mass of \(Z\) in solutions \(A\) and \(B\), and account for the difference.

Short answer

The molar mass is 128.2 g/mol in water and 259.6 g/mol in benzene. The difference is due to interactions with solvents.

Step-by-step solution

Understand the Problem

We are given two solutions of a compound \( Z \), one with water (Solution A) and one with benzene (Solution B). We need to calculate the molar mass of \( Z \) for both solutions and account for any differences.

Apply Raoult's Law

Raoult's Law states the vapor pressure of a solution is proportional to both the vapor pressure of the pure solvent and the mole fraction of the solvent. For a solvent with a non-volatile solute, the formula is \( P_{ ext{solution}} = X_{ ext{solvent}} \times P^0_{ ext{solvent}} \), where \( P^0 \) is the vapor pressure of the pure solvent.

Solve for Solution A

For Solution A (water):- The normal boiling point vapor pressure of water is \( 760 \text{ mmHg} \).- Using Raoult's Law: \( 754.5 = X_{ ext{water}} \times 760 \).- Solving for \( X_{ ext{water}} \): \( X_{ ext{water}} = \frac{754.5}{760} \approx 0.993 \).- \( X_{ ext{water}} = \frac{n_{ ext{water}}}{n_{ ext{water}} + n_{ ext{Z}}} \), where \( n \) is moles.- Moles of water: \( n_{ ext{water}} = \frac{100}{18.015} \approx 5.55 \text{ moles} \).- Reset equation to find \( n_{ ext{Z}} \): \( 0.993 = \frac{5.55}{5.55 + n_{ ext{Z}}} \).- Solve for \( n_{ ext{Z}} \): \( n_{ ext{Z}} \approx 0.039 \text{ moles} \).- Molar mass \( M = \frac{5.00}{0.039} \approx 128.2 \text{ g/mol} \).

Solve for Solution B

For Solution B (benzene):- The normal boiling point vapor pressure of benzene is \( 760 \text{ mmHg} \).- Using Raoult's Law: \( 754.5 = X_{ ext{benzene}} \times 760 \).- \( X_{ ext{benzene}} = \frac{754.5}{760} \approx 0.993 \).- \( X_{ ext{benzene}} = \frac{n_{ ext{benzene}}}{n_{ ext{benzene}} + n_{ ext{Z}}} \).- Moles of benzene: \( n_{ ext{benzene}} = \frac{100}{78.11} \approx 1.28 \text{ moles} \).- Reset equation to find \( n_{ ext{Z}} \): \( 0.993 = \frac{1.28}{1.28 + n_{ ext{Z}}} \).- Solve for \( n_{ ext{Z}} \): \( n_{ ext{Z}} \approx 0.0089 \text{ moles} \).- Molar mass \( M = \frac{2.31}{0.0089} \approx 259.6 \text{ g/mol} \).

Account for Differences

The molar mass is different for the two solutions due to different interactions between the solute \( Z \) and the solvents (water and benzene). In water, hydrogen bonding or other interactions may reduce the effective molar mass. In benzene, \( Z \) may be more isolated, showing its actual molar mass.

Key concepts

Understanding Vapor Pressure

Vapor pressure is a key concept in understanding how solutions behave. It's the pressure exerted by a vapor in equilibrium with its liquid form. In the context of Raoult's Law, the vapor pressure of a solution is less than that of its pure solvent. This is because the presence of a non-volatile solute lowers the vapor pressure.
Raoult's Law is expressed as:

• \( P_{\text{solution}} = X_{\text{solvent}} \times P^0_{\text{solvent}} \)

where \( P_{\text{solution}} \) is the vapor pressure of the solution, \( X_{\text{solvent}} \) is the mole fraction of the solvent, and \( P^0_{\text{solvent}} \) is the vapor pressure of the pure solvent.
This law illustrates that the vapor pressure of the solution decreases as the mole fraction of the solvent decreases due to the addition of solute.

Calculating Molar Mass Through Raoult's Law

When calculating molar mass using Raoult's Law, we apply the formula by measuring vapor pressure changes.
In our exercise:

• Moles of the solvent are calculated using its mass and molar mass.
• The mole fraction of the solvent is then deduced by the vapor pressure ratio.
• Finally, solve for the moles of the solute and then the molar mass: \( M = \frac{\text{mass of solute}}{\text{moles of solute}} \)

This approach leverages vapor pressure measurements to find the molar mass by comparing the solution's behavior against its pure solvent's properties.

Influence of Solute-Solvent Interactions

Different solutes and solvents interact uniquely, greatly affecting solution behavior. These interactions can explain discrepancies in molar mass calculations.
In water, solute interactions such as hydrogen bonding can lead to a lower effective concentration, affecting vapor pressure. In benzene, where there may be fewer interactions, the solute's effect is more reflective of its true molar mass.

• This highlights how intermolecular forces play a crucial role in solutions.
• Hydrogen bonding and van der Waals interactions are examples that can shift expected outcomes.

Understanding these interactions is vital in predicting the properties of solutions.