Question

Some nonelectrolyte solute (molar mass \(=142 \mathrm{g} / \mathrm{mol}\) ) was dissolved in \(150 . \mathrm{mL}\) of a solvent (density \(=0.879 \mathrm{g} / \mathrm{cm}^{3}\) ). The elevated boiling point of the solution was \(355.4 \mathrm{K} .\) What mass of solute was dissolved in the solvent? For the solvent, the enthalpy of vaporization is \(33.90 \mathrm{kJ} / \mathrm{mol},\) the entropy of vaporization is \(95.95 \mathrm{J} / \mathrm{K} \cdot \mathrm{mol},\) and the boiling-point elevation constant is \(2.5 \mathrm{K} \cdot \mathrm{kg} / \mathrm{mol}.\)

Short answer

The mass of the solute dissolved in the solvent is approximately 410 g.

Step-by-step solution

Calculate the mass of the solvent

The density of the solvent is given as 0.879 g/cm³ and the volume is 150 mL. Using the conversion factor 1 mL = 1 cm³, multiply the volume of the solvent by its density to find the mass of the solvent: Mass of solvent = Density of solvent * Volume of solvent = \(0.879 \, g/cm^3 * 150 \, mL = 131.85 \, g\)

Calculate the change in temperature, ΔT

The boiling point of the pure solvent is not provided, but it can be calculated using the Clausius-Clapeyron equation: \(\Delta H_{vap} = \Delta S_{vap} * (T_{b(solvent)} - T_{b(solution)})\) Given that \(\Delta H_{vap} = 33.90 \, kJ/mol = 33900 \, J/mol\) and \(\Delta S_{vap} = 95.95 \, J/(K \cdot mol)\), we can rearrange the equation above to solve for the boiling point of the pure solvent, \(T_{b(solvent)}\): \(T_{b(solvent)} = T_{b(solution)} + \frac{\Delta H_{vap}}{\Delta S_{vap}} = 355.4 \, K + \frac{33900 \, J/mol}{95.95 \, J/(K \cdot mol)} = 410.16 \, K\) Subtract the elevated boiling point from the boiling point of the pure solvent and we get: \(\Delta T = T_{b(solvent)} - T_{b(solution)} = 410.16\, K - 355.4\, K = 54.76\, K\)

Calculate the moles of solute, n

Using the formula for boiling point elevation, \(\Delta T_{b} = K_{b} \cdot m\), where \(K_{b}\) is the boiling-point elevation constant and \(m\) is the molality, rearrange the equation and solve for the moles of solute, \(n\): \(m = \frac{\Delta T_{b}}{K_{b}} = \frac{54.76\, K}{2.5\, K\cdot kg/mol} = 21.9\, mol/kg = 0.0219\, mol/g\) Molality is defined as moles of solute per kilogram of solvent. Rearrange the equation \(m = \frac{n}{mass_{solvent}}\) to solve for n and get: \(n = m \cdot mass_{solvent} = 0.0219\, mol/g \cdot 131.85\, g = 2.886\, mol\)

Calculate the mass of the solute

Lastly, multiply the number of moles by the molar mass of the solute to get the mass of the solute: Mass of solute = Molar mass of solute * n = 142\, g/mol * 2.886\, mol = 409.8\, g So, approximately 410 g of solute was dissolved in the solvent.

Key concepts

Colligative Properties

Colligative properties are those physical properties of solutions that depend on the number of dissolved particles in a given amount of solvent, and not on the nature of those particles. This means that whether you dissolve salt, sugar, or any other solute in water, as long as the number of particles is the same, the impact on the solution's colligative properties will be similar.

Critical colligative properties include vapor pressure lowering, freezing point depression, boiling point elevation, and osmotic pressure. Boiling point elevation, in particular, is when a solute is added to a solvent, resulting in a solution that boils at a higher temperature than the pure solvent. This phenomenon occurs because the presence of solute particles hinders the ability of solvent molecules to escape into the gas phase.

To better understand the amount of solute present in a solution, we often use measurements like molality, which plays a crucial role in quantifying changes to colligative properties.

Molality

Molality (\(m\)) is a measure of solute concentration in a solution. It is defined as the moles of solute divided by the kilograms of solvent. Unlike molarity, which depends on the volume of the solution, molality depends solely on the mass, which is not affected by temperature changes. This makes molality a more precise measurement for use in colligative properties and temperature-related problems. To calculate the molality, you take the number of moles of solute and divide it by the mass of the solvent in kilograms.

For instance, in the problem at hand, the molality is crucial for calculating the boiling point elevation. After finding the mass of the solvent to be about 131.85 grams, we convert it to kilograms (0.13185 kg) before using it to find the moles of solute, which helps us understand the extent to which the boiling point is elevated.

Clausius-Clapeyron Equation

The Clausius-Clapeyron equation provides a way to relate the vapor pressure of a liquid to its temperature and thus calculate the boiling point under different circumstances. It's an essential part of thermodynamics and physical chemistry, connecting the enthalpy of vaporization (\(\Delta H_{vap}\)) and the entropy of vaporization (\(\Delta S_{vap}\)) to the temperature at which a liquid transforms into a gas.

In our case, this equation allows us to back-calculate the original boiling point of the solvent, using the enthalpy and entropy of vaporization given for the pure solvent. Knowing the original boiling point and the elevation due to the solute gives us the increase in temperature (\(\Delta T\)), which directly connects to the molality of the solution. This is a classic demonstration of how thermodynamic principles can be applied to practical problems, such as determining the amount of a solute in a boiling-point elevation scenario.