Problem 79 - Chemistry The Molecular Nature of Matter

Question

A solution containing \(8.3 \mathrm{~g}\) of a nonvolatile, nondissociating substance dissolved in \(1.00 \mathrm{~mol}\) of chloroform, \(\mathrm{CHCl}_{3},\) has a vapor pressure of 511 torr. The vapor pressure of pure \(\mathrm{CHCl}_{3}\) at the same temperature is 526 torr. Calculate (a) the mole fraction of the solute, (b) the number of moles of solute in the solution, and (c) the molecular mass of the solute.

Step-by-Step Solution

Answer

The mole fraction of the solute is 0.0285, the number of moles of solute in the solution is 0.0296 moles, and the molecular mass of the solute is 280.41 g/mol.

Step 1: Calculate the Mole Fraction of the Solute

Use Raoult's Law which states that the vapor pressure of the solution is proportional to the mole fraction of the solvent. Calculate the mole fraction of the solute, \( chi_{solute} \), by using the formula \( P = P^{\text{*}} - \Delta P \) where \( P^{\text{*}} \) is the vapor pressure of the pure solvent and \( P \) is the vapor pressure of the solution. The vapor pressure lowering, \( \Delta P = P^{\text{*}} - P \). Substitute the given values \( \Delta P = 526 \text{ torr} - 511 \text{ torr} = 15 \text{ torr} \). The mole fraction of the solute is then \( chi_{solute} = \frac{\Delta P }{P^{\text{*}}} = \frac{15 \text{ torr}}{526 \text{ torr}} \) .

Step 2: Calculate the Number of Moles of the Solute in the Solution

Since we have the mole fraction of the solute, and we know there is 1 mole of chloroform, we can find the number of moles of the solute. The mole fraction formula is \( chi_{solute} = \frac{n_{solute}}{n_{solute} + n_{solvent}} \), and solving for \( n_{solute} \) gives us \( n_{solute} = chi_{solute} \times \left(n_{solvent} + n_{solute}\right) - chi_{solute} \times n_{solute} \). Since chloroform is the solvent, \( n_{solvent} = 1 \) mole, we can rearrange to get \( n_{solute} = \frac{chi_{solute}}{1 - chi_{solute}} \) and then plug in the value of \( chi_{solute} \) calculated in Step 1.

Step 3: Calculate the Molecular Mass of the Solute

With the mass of the solute and the number of moles of the solute, we can now find the molecular mass (molar mass) of the solute by using the formula \( MM = \frac{m}{n} \) where \( MM \) is the molecular mass, \( m \) is the mass of the solute in grams, and \( n \) is the number of moles of the solute. Substitute \( m = 8.3 \text{ g} \) and the value of \( n_{solute} \) from Step 2 to find \( MM \) .

Key Concepts

Mole Fraction Calculation

Understanding the concept of mole fraction is critical for comprehending various phenomena in chemistry, such as colligative properties and vapor pressure. A mole fraction indicates the proportion of a component within a mixture and is calculated without units, making it a ratio of moles to the total moles present.

For example, if a solution is composed of solute and solvent, the mole fraction of each can be expressed as follows:

• Mole fraction of solute (\(\chi_{solute}\)): \(\chi_{solute} = \frac{n_{solute}}{n_{solute} + n_{solvent}}\), where \(n\) represents the number of moles.
• Mole fraction of solvent (\(\chi_{solvent}\)): Similarly, this can be calculated using the moles of solvent.

To find the mole fraction from the vapor pressure data, you can use Raoult's Law. Raoult's Law connects vapor pressure with the mole fraction of a solvent in a solution. If the vapor pressure of a solvent is lowered when a solute is dissolved in it, this drop in vapor pressure can be used to calculate the mole fraction of the solute.
The underlying principle is that the presence of a nonvolatile solute reduces the number of solvent particles that can escape into the vapor phase, effecting a decrease in vapor pressure. With the provided values for pure chloroform vapor pressure and the mixture's vapor pressure, you can determine the mole fraction of the solute, which is integral for further calculations such as determining the number of moles and molecular mass of the solute.

Molecular Mass Determination

Determining molecular mass, sometimes referred to as molar mass, is a basic task in chemistry that enables you to convert between the mass of a substance and the amount in moles. The molecular mass indicates how much one mole of a substance weighs, and it is measured in grams per mole (g/mol).

The general formula to determine molecular mass (\( MM \)) is:\[ MM = \frac{m}{n} \]Where:

• \( m \) is the mass of the solute, in grams.
• \( n \) is the number of moles of the solute.

Using the mass of the solute given in the problem and the number of moles calculated from the mole fraction, one can find the molecular mass of the solute. This is crucial for identifying the compound or comparing its molecular mass to those of known substances. Molecular mass determination directly impacts the understanding of the chemical composition of molecules and their interactions with each other in mixtures or reactions.

Nonvolatile Solute in Solution

A nonvolatile solute is a substance that does not readily evaporate, or turn into gas, when dissolved in a solution. This property significantly impacts the colligative properties of the solution, such as boiling point elevation, freezing point depression, and vapor pressure lowering. The nonvolatile nature means that the solute particles do not contribute to the vapor pressure; only the solvent does because it's the volatile component that can vaporize.

The impact of a nonvolatile solute on vapor pressure is described by Raoult's Law, which relates the drop in vapor pressure to the presence of the solute. Following this law, when a nonvolatile solute is dissolved in a solvent, the vapor pressure of the resulting solution is less than that of the pure solvent. This occurs because the solute particles occupy space at the surface of the solution, thereby reducing the number of solvent molecules that can escape into the vapor phase.
Understanding how a nonvolatile solute behaves is essential for predicting how a solution will react under various conditions, including those pertinent to industrial processes, laboratory experiments, and even in cooking recipes where syrup concentration is important.