Problem 100 - Chemistry The Central Science

Question

A solution contains 0.115 \(\mathrm{mol} \mathrm{H}_{2} \mathrm{O}\) and an unknown number of moles of sodium chloride. The vapor pressure of the solution at \(30^{\circ} \mathrm{C}\) is 25.7 torr. The vapor pressure of pure water at this temperature is 31.8 torr. Calculate the number of grams of sodium chloride in the solution. (Hint: Remember that sodium chloride is a strong electrolyte.)

Step-by-Step Solution

Answer

The mass of sodium chloride in the solution is 3.20 grams.

Step 1: Use Raoult's law to find the mole fraction of water

According to Raoult's law, the vapor pressure of a component in a solution is equal to the product of the mole fraction and the vapor pressure of the pure component. \(P = P_A * X_A\) Here, \(P\) is the vapor pressure of the solution, \(P_A\) is the vapor pressure of the pure component (water), and \(X_A\) is the mole fraction of the component (water) in the solution. We are given \(P = 25.7\) torr, \(P_A = 31.8\) torr, and we need to find the mole fraction of water, \(X_A\). \( X_A = \frac{P}{P_A} \)

Step 2: Calculate the mole fraction of water

Plug in the values of vapor pressures into the formula and calculate the mole fraction of water: \( X_A = \frac{25.7}{31.8} \) \( X_A = 0.808\) Since the mole fraction of a solute and solvent sum up to 1, the mole fraction of sodium chloride is: \(X_{NaCl} = 1 - X_A = 1 - 0.808 = 0.192\)

Step 3: Calculate the moles of sodium chloride

Now that we know the mole fraction of sodium chloride, we can calculate the moles of sodium chloride based on the moles of water: Moles of \(H_2O = 0.115\) Since we are given the mole fractions of water and sodium chloride: Moles of \(NaCl = \frac{X_{NaCl}}{X_A} \times\) Moles of \(H_2O\) \(= \frac{0.192}{0.808} \times 0.115\) Since sodium chloride is a strong electrolyte, it dissociates completely into its ions: \(Na^+\) and \(Cl^-\), and we need to double the moles of sodium chloride to get the total moles of ions: Moles of ions of \(NaCl = 2 \times\) Moles of \(NaCl\) Moles of ions of \(NaCl = 2 \times \frac{0.192}{0.808} \times 0.115\) Moles of ions of \(NaCl = 0.0548\)

Step 4: Calculate the mass of sodium chloride

Finally, we can calculate the mass of sodium chloride using the moles and molar mass. Molar mass of \(NaCl = 58.44 \frac{g}{mol}\) Mass of \(NaCl =\) Moles of ions of \(NaCl \times\) Molar mass of \(NaCl\) Mass of \(NaCl = 0.0548 \times 58.44 g/mol\) Mass of \(NaCl = 3.20 g\) So, the mass of sodium chloride in the solution is 3.20 grams.

Key Concepts

Raoult's Law

This principle plays a crucial role in determining the vapor pressure of a solution. Raoult's law states that the partial vapor pressure of each component in an ideal solution is directly proportional to its mole fraction.

In mathematical terms, the law is expressed as:
\( P = P^{\ast} \cdot X \)
Here,

• \( P \) is the vapor pressure of the solute in the solution,
• \( P^{\ast} \) is the vapor pressure of the pure solvent, and
• \( X \) is the mole fraction of the solvent in the solution.

To understand how the law translates to practical scenarios, think of it as a way to calculate the diminished vapor pressure due to the presence of a solute. In a solution, the solute particles occupy space at the surface, where evaporation occurs, which reduces the number of solvent molecules that can escape into the vapor phase. As a result, the vapor pressure is lower compared to that of the pure solvent. This understanding is essential when solving problems involving vapor pressure and composition of solutions, such as calculating the mass of a solute like sodium chloride in a solution.

Mole Fraction

The mole fraction is a way of expressing the composition of a component within a mixture or solution. It is defined as the ratio of the moles of a particular component to the total moles of all the components present.

To calculate it, use the formula:
\( X_i = \frac{n_i}{n_{total}} \)
where

• \( X_i \) is the mole fraction,
• \( n_i \) is the number of moles of the component, and
• \( n_{total} \) is the total moles in the mixture.

Understanding the mole fraction is essential for various calculations in chemistry, including using Raoult's law to find the vapor pressure of solutions. It is especially important when dealing with mixtures where the number of moles of each component affects the overall properties of the solution.

Sodium Chloride Solubility

When discussing sodium chloride solubility, you're talking about how much of this salt can dissolve in water. Sodium chloride, common table salt, is known for its high solubility in water.

However, it's important to note that solubility isn't infinite. It has a limit, varying with temperature and other factors, after which additional salt will not dissolve. This characteristic influences calculations involving the concentration of solutions. In the context of vapor pressure, solubility is tied to the solute's ability to disrupt the solvent's surface and hence lower the vapor pressure according to the principles of Raoult's law.

Sodium chloride is also a strong electrolyte, meaning that it completely disassociates into its ions when dissolved in water. This complete ionization must be considered when calculating properties like mole fraction and vapor pressure, as it affects the number of particles interacting in the solution.

Strong Electrolyte

A strong electrolyte is a substance that completely dissociates into ions when dissolved in a solvent like water. This is an essential concept in chemistry, especially in the discussion of ionic compounds like sodium chloride (NaCl).

When NaCl dissolves, it separates into sodium ions (Na+) and chloride ions (Cl-), which can conduct electricity. The full disassociation implies that for every mole of NaCl dissolved, there are actually two moles of ions contributing to the properties of the solution. This fact is vitally important when using equations like Raoult's law or calculating mole fractions, as failing to account for the doubled number of particles can lead to significant errors in the calculated properties of the solution.

Hence, when working through problems that involve strong electrolytes, one must adjust calculations to reflect the increase in particles due to the disassociation process.