Problem 35 - Chemistry: Principles and Reactions

Question

The vapor pressure of pure \(\mathrm{CCl}_{4}\) at \(65^{\circ} \mathrm{C}\) is \(504 \mathrm{~mm} \mathrm{Hg}\). How many grams of naphthalene \(\left(\mathrm{C}_{10} \mathrm{H}_{8}\right)\) must be added to \(25.00 \mathrm{~g}\) of \(\mathrm{CCl}_{4}\) so that the vapor pressure of \(\mathrm{CCl}_{4}\) over the solution is \(483 \mathrm{~mm} \mathrm{Hg}\) ? Assume the vapor pressure of naphthalene at \(65^{\circ} \mathrm{C}\) is negligible.

Step-by-Step Solution

Answer

Answer: Approximately 5.13 grams of naphthalene should be added.

Step 1: Write the given information

The given information can be written as follows: - The vapor pressure of pure \(\mathrm{CCl}_{4}\) (\(P_{A}^{*}\)) is \(504 \mathrm{~mm} \mathrm{Hg}\). - The mass of \(\mathrm{CCl}_{4}\) is \(25.00~\mathrm{g}\). - The vapor pressure of the solution containing \(\mathrm{CCl}_{4}\) and naphthalene (\(P_A\)) is \(483 \mathrm{~mm} \mathrm{Hg}\). We want to find how many grams of naphthalene must be added to this solution.

Step 2: Determine the molecular weights of the substances

Before we continue, let's find the molecular weights of \(\mathrm{CCl}_{4}\) and naphthalene. The molecular weight of \(\mathrm{CCl}_{4}\): $$ 1\text{C} + 4\text{Cl} = 12.01 + 4(35.45) = 153.81\, \mathrm{g/mol} $$ The molecular weight of naphthalene: $$ 10\text{C} + 8\text{H} = 10(12.01) + 8(1.008) = 128.17\, \mathrm{g/mol} $$

Step 3: Apply Raoult's Law and mole fraction

According to Raoult's Law, the partial pressure of a component in a solution is proportional to its mole fraction: $$ P_A = x_A P_A^{*} $$ where \(P_A\) is the vapor pressure of \(\mathrm{CCl}_{4}\) over the solution, \(x_A\) is the mole fraction of \(\mathrm{CCl}_{4}\) and \(P_A^{*}\) is the vapor pressure of pure \(\mathrm{CCl}_{4}\). We can calculate the mole fraction of \(\mathrm{CCl}_{4}\) as follows: $$ x_A = \frac{n_A}{n_A + n_B} $$ where \(n_A\) is the number of moles of \(\mathrm{CCl}_{4}\) and \(n_B\) is the number of moles of naphthalene.

Step 4: Calculate the number of moles of both substances

Now, let's find the number of moles of \(\mathrm{CCl}_{4}\) and naphthalene. Number of moles of \(\mathrm{CCl}_{4}\): $$ n_A = \frac{25.00\, \mathrm{g}}{153.81\, \mathrm{g/mol}} \approx 0.1624~\mathrm{mol} $$ Let's denote the mass of naphthalene as \(m_B\) in grams. The number of moles of naphthalene is as follows: $$ n_B = \frac{m_B}{128.17\, \mathrm{g/mol}} $$

Step 5: Substitute the values into the mole fraction equation

Plugging the values of \(n_A\) and \(n_B\) into the mole fraction equation: $$ x_A = \frac{0.1624}{0.1624 + \frac{m_B}{128.17}} $$

Step 6: Substitute the values into Raoult's Law equation

Now, using the mole fraction of \(\mathrm{CCl}_{4}\), we can plug the values into Raoult's Law equation: $$ 483 = \frac{0.1624}{0.1624 + \frac{m_B}{128.17}}\cdot 504 $$

Step 7: Solve for the mass of naphthalene (\(m_B\))

Finally, let's solve this equation to find the mass of naphthalene: $$ m_B = \frac{128.17\cdot 0.1624 (504 - 483)}{(483)(0.1624)} \approx 5.13\, \mathrm{g} $$ The mass of naphthalene that should be added to \(25.00\, \mathrm{g}\) of \(\mathrm{CCl}_{4}\) to obtain a vapor pressure of \(483\, \mathrm{mm} \mathrm{Hg}\) is approximately \(5.13\, \mathrm{g}\).

Key Concepts

Vapor Pressure

Vapor pressure is the pressure exerted by the vapor of a liquid when it is in equilibrium with its liquid phase. This means at a given temperature, the liquid and its vapor coexist in balance. Generally, the vapor pressure increases with temperature because more molecules have enough energy to escape into the vapor phase.

In the context of Raoult's Law, the vapor pressure of a solvent will decrease when a solute is added, which happens due to the interaction between solute and solvent molecules. Consider pure carbon tetrachloride (CCl extsubscript{4}), which has a vapor pressure of 504 mmHg at 65°C. When naphthalene, whose own vapor pressure is negligible at this temperature, is added, the overall vapor pressure of CCl extsubscript{4} in the solution drops to 483 mmHg. This decrease is explained by the fact that fewer CCl extsubscript{4} molecules will escape into the vapor phase when they are mixed with naphthalene molecules.

Understanding vapor pressure is essential for predicting how adding substances to a solvent can change the equilibrium between the liquid and gas phases, which is especially relevant in chemistry and solutions.

Mole Fraction

The mole fraction is a way of expressing the concentration of a component in a mixture or solution. It is denoted as the ratio of the number of moles of that component to the total number of moles of all components in the solution. This is mathematically expressed as:

• For component A: \( x_A = \frac{n_A}{n_A + n_B} \)

Here, \( n_A \) represents the moles of CCl extsubscript{4}, while \( n_B \) symbolizes the moles of naphthalene.

In Raoult's Law, the mole fraction plays a critical role in determining the partial vapor pressure of the solvent. If you know the mole fraction of a solvent in a solution, you can multiply it by the vapor pressure of the pure solvent to find the solution's vapor pressure.

For example, in the solution described, if CCl extsubscript{4} is the solvent, then calculating its mole fraction helps us determine that its reduced vapor pressure, based on the equation \( P_A = x_A P_A^{*} \), will influence the molecular behavior of the solution.

Molecular Weight

Molecular weight, also known as molecular mass, is the sum of the atomic weights of all atoms in a molecule. It is expressed in grams per mole (g/mol) and is critical in quantifying the number of moles in a given mass of a substance.

For practical applications like the problem above, knowing the molecular weight of CCl extsubscript{4} (153.81 g/mol) and naphthalene (128.17 g/mol) allows us to convert between mass and moles, which is necessary for calculations involving mole fractions and Raoult's Law.

In the solution, to find how many grams of naphthalene must be added to the solution for a certain vapor pressure, we used molecular weights to find the moles of both CCl extsubscript{4} and naphthalene. With \( n_A = \frac{25.00 \, ext{g}}{153.81 \, ext{g/mol}} \approx 0.1624 \, ext{mol} \), the moles of naphthalene are calculated similarly based on an arbitrary mass \( m_B \), leading to the correct solution.