Problem 100 - Chemistry The Central Science In Si Units

Question

A solution contains \(0.50 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}\) and an unknown number of moles of sodium chloride. The vapor pressure of the solution at \(29^{\circ} \mathrm{C}\) is \(3.85 \mathrm{kPa}\). The vapor pressure of pure water at this temperature is \(4.05 \mathrm{kPa}\). Calculate the number of grams of sodium chloride in the solution.

Step-by-Step Solution

Answer

The solution contains approximately 1.51 grams of sodium chloride.

Step 1: Understand Raoult's Law

Raoult's Law is used to describe the vapor pressure of a solution. According to Raoult's Law, the vapor pressure of the solution is equal to the mole fraction of the solvent times the vapor pressure of the pure solvent: \( P_{ ext{solution}} = X_{ ext{solvent}} \times P^0_{ ext{solvent}} \). Here, \( P_{ ext{solution}} = 3.85 \text{ kPa} \) and \( P^0_{ ext{solvent}} = 4.05 \text{ kPa} \).

Step 2: Express Mole Fraction

The mole fraction (\(X_{ ext{solvent}}\)) is the ratio of moles of the solvent to the total moles in the solution. This can be written as \(X_{ ext{solvent}} = \frac{n_{ ext{H}_2 ext{O}}}{n_{ ext{H}_2 ext{O}} + n_{ ext{NaCl}}} \), where \(n_{ ext{H}_2 ext{O}} = 0.50 \text{ mol}\).

Step 3: Solve for Mole Fraction

Using Raoult's Law, substitute the known values to find the mole fraction: \( 3.85 = X_{ ext{H}_2 ext{O}} \times 4.05 \). Solve for \(X_{ ext{H}_2 ext{O}}\): \(X_{ ext{H}_2 ext{O}} = \frac{3.85}{4.05} \approx 0.9506\).

Step 4: Substitute Mole Fraction and Solve for \(n_{\text{NaCl}}\)

Substitute the calculated mole fraction back into the equation: \(0.9506 = \frac{0.50}{0.50 + n_{\text{NaCl}}}\). Solve for \(n_{\text{NaCl}}\):\\[ 0.9506 (0.50 + n_{\text{NaCl}}) = 0.50 \]\\[ 0.4753 + 0.9506 n_{\text{NaCl}} = 0.50 \]\\[ 0.9506 n_{\text{NaCl}} = 0.0247 \]\\[ n_{\text{NaCl}} = \frac{0.0247}{0.9506} \approx 0.0259 \text{ mol} \].

Step 5: Convert Moles of NaCl to Grams

To find the mass of sodium chloride, use its molar mass (\(58.44 \text{ g/mol}\)). Multiply the number of moles by the molar mass: \( 0.0259 \times 58.44 \approx 1.51 \text{ g}\).

Key Concepts

Vapor Pressure

Vapor pressure is a fascinating concept in the world of chemistry, especially when we talk about solutions. Think of vapor pressure as the pressure exerted by a vapor in equilibrium with its liquid form. Imagine a closed container where a liquid is slowly turning into vapor and then back to liquid. At some point, the number of molecules leaving the liquid and entering the vapor state equals the number returning to the liquid. This balance creates a specific pressure, known as vapor pressure.
In the context of Raoult's Law, vapor pressure helps us understand how adding a non-volatile solute (like sodium chloride) affects the pressure of the solvent's vapor. When we dissolve something like sodium chloride in water, fewer water molecules are available at the surface to evaporate, which decreases the vapor pressure compared to pure water. This decrease in vapor pressure due to the presence of a solute tells us something important about the solution’s composition.

Mole Fraction

The mole fraction is a very handy way to express the composition of a solution. It tells us the proportion of one component compared to the entire mixture. For example, in our water and sodium chloride solution, the mole fraction of water would be the number of moles of water divided by the total number of moles of all components in the solution.
Mathematically, this is shown as:

• \[ X_{\text{solvent}} = \frac{n_{\text{H}_2\text{O}}}{n_{\text{H}_2\text{O}} + n_{\text{NaCl}}} \]

Here, the mole fraction helps us apply Raoult's Law effectively. By knowing the mole fraction of water, we can determine how much the addition of sodium chloride has lowered the vapor pressure from that of pure water. This calculation is a crucial part of understanding how solutes like sodium chloride can change the vapor pressure of solvents they are dissolved in.

Sodium Chloride Calculation

Calculating the number of moles and subsequently the mass of sodium chloride in a solution involves using both Raoult’s Law and the concept of the mole fraction. Once we determine the mole fraction of the solvent (here, water), we can find out how many moles of sodium chloride are present.
Starting from Raoult’s Law, we substitute known values to find the mole fraction of water. Upon solving the equation

• \[ 3.85 = X_{\text{H}_2\text{O}} \times 4.05 \]
• \[ X_{\text{H}_2\text{O}} = \frac{3.85}{4.05} \approx 0.9506 \]

we get the value for the water mole fraction, which we then use to find the amount of sodium chloride. With water at 0.50 moles, solving for sodium chloride involves rearranging and substituting back into the total moles equation to find Sodium Chloride's moles:

• \[ 0.9506 = \frac{0.50}{0.50 + n_{\text{NaCl}}} \]

Once the moles of sodium chloride are known, converting it to grams involves multiplying by its molar mass (58.44 g/mol). It’s crucial to keep these steps orderly to accurately determine the mass of sodium chloride, which in this case is around 1.51 grams.