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 \(\mathrm{J} / \mathrm{K} \cdot\) 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 \(16.79\ \mathrm{g}\).

Step-by-step solution

Calculate the boiling-point elevation

The boiling-point elevation can be calculated using the formula: \(\Delta T_b = K_b \cdot m\) where \(\Delta T_b\) is the boiling-point elevation, \(K_b\) is the boiling-point elevation constant, and \(m\) is the molality of the solute in the solution. We are given the boiling-point elevation constant, \(K_b = 2.5\ \mathrm{K} \cdot \mathrm{kg} / \mathrm{mol}\), and the boiling point of the solution, \(T_{solution} = 355.4\ \mathrm{K}\). First, we find the normal boiling point of the solvent, \(T_{solvent}\), using the given values of enthalpy and entropy of vaporization: \(T_{solvent}=\frac{\Delta H_v}{\Delta S_v}\) where \(\Delta H_v = 33.90\ \mathrm{kJ} / \mathrm{mol} = 33900\ \mathrm{J} / \mathrm{mol}\) and \(\Delta S_v = 95.95\ \mathrm{J} / \mathrm{K} \cdot \mathrm{mol}\). \(T_{solvent}=\frac{33900\ \mathrm{J} / \mathrm{mol}}{95.95\ \mathrm{J} / \mathrm{K} \cdot \mathrm{mol}} = 353.16\ \mathrm{K}\) Now, we can find the boiling-point elevation: \(\Delta T_b = T_{solution} - T_{solvent} = 355.4\ \mathrm{K} - 353.16\ \mathrm{K} = 2.24\ \mathrm{K}\)

Calculate the molality of the solute

Now that we have the boiling-point elevation, we can calculate the molality of the solute using the formula: \(m=\frac{\Delta T_b}{K_b}\) \(m=\frac{2.24\ \mathrm{K}}{2.5\ \mathrm{K} \cdot \mathrm{kg} / \mathrm{mol}}=0.896 \ \mathrm{mol} / \mathrm{kg}\)

Calculate the mass of the solvent

To calculate the mass of the solvent, we use the given volume and density of the solvent. Mass of solvent = Volume × Density where Volume = \(150\ \mathrm{mL} = 0.150\ \mathrm{L}\) and Density = \(0.879\ \mathrm{g} / \mathrm{cm}^{3}\). To convert volume from liters to cubic centimeters, we can multiply by 1000. Volume in cubic centimeters = \(0.150\ \mathrm{L} \times 1000 = 150\ \mathrm{cm}^{3}\) Now, we can find the mass of the solvent: Mass of solvent = \(150\ \mathrm{cm}^{3} \times 0.879\ \mathrm{g} / \mathrm{cm}^{3} = 131.85\ \mathrm{g}\)

Calculate the mass of the solute

Now that we know the molality of the solute and the mass of the solvent, we can find the moles of solute using the formula: moles of solute = \(\frac{m \times \text{Mass of solvent}}{\text{Molar mass of solvent}}\) Moles of solute = \(\frac{0.896\ \mathrm{mol} / \mathrm{kg} \times 131.85\ \mathrm{g}}{1000\ \mathrm{g}/\mathrm{kg}} = 0.1182\ \mathrm{mol}\) Finally, we find the mass of the solute using the given molar mass of the solute, \(M = 142\ \mathrm{g} / \mathrm{mol}\). Mass of solute = moles of solute × molar mass of solute Mass of solute = \(0.1182\ \mathrm{mol} \times 142\ \mathrm{g} / \mathrm{mol} = 16.79\ \mathrm{g}\) So, the mass of solute dissolved in the solvent is 16.79 g.

Key concepts

Molality

Molality is a measure of the concentration of a solute in a solution. It's calculated based on the amount of solute and the mass of the solvent, rather than the total volume of the solution. This is useful because volume can change with temperature, but mass does not. Molality (m) is defined as:

• The number of moles of solute per kilogram of solvent

We often express molality in \( \ ext{mol/kg} \). To find it, you divide the moles of solute by the mass of the solvent in kilograms. In the context of our exercise, once the boiling-point elevation is known, we used the formula:

• \( m = \frac{\Delta T_b}{K_b} \)

By inserting the values, students can see how molality relates to changes in boiling points, helping to understand solution behaviors better.

Enthalpy of Vaporization

Enthalpy of vaporization is a measure of the energy required to vaporize a liquid. It tells us how much heat is needed to convert one mole of a liquid into vapor at constant pressure. The formula involves:

• \( \Delta H_v \) which is given in \( \text{kJ/mol} \)

During phase changes, this value represents the energy investment required to break the molecular interactions holding the liquid together. In our specific case, the enthalpy of vaporization assisted in determining the normal boiling point of the solvent by using it in conjunction with the entropy of vaporization value.

Entropy of Vaporization

Entropy of vaporization quantifies the disorder introduced when a liquid turns into vapor. It represents the increase in randomness or dispersal of matter and energy. It can be expressed in \( \text{J/K} \cdot \text{mol} \).This value is crucial because changes in entropy provide insight into the feasibility and spontaneity of phase transitions. In the exercise, by dividing the enthalpy of vaporization by the entropy, we determined the normal boiling point of the solvent. This value was then used to calculate the boiling-point elevation, illustrating the relationship between thermodynamic properties and physical changes.

Nonelectrolyte Solution

An understanding of nonelectrolyte solutions is key when discussing boiling-point elevation. These solutions contain solutes that do not dissociate into ions when dissolved in a solvent. As a result:

• They don’t conduct electricity
• They aren’t associated with any ionization-related properties

This feature simplifies the calculation of boiling or freezing point changes, as each molecule of solute acts as a discrete particle in the solution without contributing to the total ion count. In this exercise, the solute fit this description, allowing the step-by-step method using basic colligative property principles to determine changes in boiling points due to the presence of the solute.