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 \(18.0\, \mathrm{g}\).

Step-by-step solution

Calculate the mass of the solvent

Since we know the volume of the solvent (\(150\, \mathrm{mL}\)) and the density of the solvent (\(0.879\, \mathrm{g/cm^3}\)), we can first calculate the total mass of the solvent by multiplying the volume and the density: \(\text{mass of solvent}=150\, \mathrm{mL}\times 0.879\, \mathrm{g/cm^3}=131.85\, \mathrm{g}\)

Find the normal boiling point of the solvent

We need to find the normal boiling point of the solvent. We have the entropy and enthalpy of vaporization, so we can use the Clausius-Clapeyron equation: \(\ln{\frac{P_2}{P_1}}=-\frac{\Delta{H_{vap}}}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)\) Since boiling takes place at the temperature where vapor pressure equals atmospheric pressure, we can assume that \(P_1=P_2=1 \,\text{atm}\). With this assumption, the equation becomes: \(0=-\frac{\Delta{H_{vap}}}{R}\left(\frac{1}{T_b}-\frac{1}{T_1}\right)\) Rearranging the equation, we find: \(T_b =\frac{\Delta{H_{vap}}}{\Delta{S_{vap}}}=\frac{33.90\, \mathrm{kJ/mol}}{95.95\, \mathrm{J/K}\cdot \mathrm{mol}}=353\, \mathrm{K}\)

Calculate the molality of the solution

We will now use the boiling point elevation formula to find the molality of the solution. The formula for boiling point elevation is: \(\Delta{T_b}=K_b \cdot m\) We know the boiling point elevation constant, \(K_b=2.5\, \mathrm{K⋅kg/mol}\) and the elevated boiling point is \(355.4\, \mathrm{K}\), so we can find the molality, \(m\), as follows: \(m=\frac{\Delta{T_b}}{K_b}=\frac{355.4 - 353}{2.5}=\frac{2.4}{2.5}=0.96\, \mathrm{mol/kg}\)

Calculate the mass of the solute

Now that we have the molality of the solution, we can find the mass of the solute. We can use the relationship between molality, moles of solute, mass of solute and mass of solvent to write: \(m=\frac{\text{moles of solute}}{\text{mass of solvent (kg)}}\) Since we now know the molality, we can rearrange this equation to find the moles of solute: \(\text{moles of solute}= m\times \text{mass of solvent (kg)}=0.96\,\mathrm{mol/kg} \times \frac{131.85\, \mathrm{g}}{1000\, \mathrm{g/kg}}=0.1267\, \mathrm{mol}\) We can compute the mass of solute using the molar mass of the solute (\(142\, \mathrm{g/mol}\)): \(\text{mass of solute}=\text{moles of solute}\times \text{molar mass of solute}=0.1267\, \mathrm{mol}\times 142\, \mathrm{g/mol}=18.0\, \mathrm{g}\) So, the mass of the solute dissolved in the solvent is approximately \(18.0\, \mathrm{g}\).

Key concepts

Boiling Point Elevation

Understanding boiling point elevation is key for students who are tackling problems related to changes in a solution's boiling point after adding a solute. This phenomenon occurs due to the presence of solute particles, which disrupt the solvent's ability to enter the vapor phase at the normal boiling temperature. As a result, more energy (in the form of heat) is required for the boiling to occur, hence the boiling point elevates.

Boiling point elevation is a colligative property, meaning it is dependent on the number of solute particles in the solvent, not the type of particles. The equation describing this relationship is quite simple: \(\text{Boiling point elevation (}\text{Δ}T_b\text{)} = K_b \times m\), where \(K_b\) is the ebullioscopic constant and \(m\) represents the molality of the solution. In the context of the exercise, we leveraged this equation to derive the molality of the solution using the observed boiling point elevation.

Molality

Delving into the concept of molality, it's important for students to grasp this measure of concentration of a solution. Molality (\(m\)) is defined as the moles of solute per kilogram of solvent, not to be confused with molarity, which is moles of solute per liter of solution. The distinction is significant because molality is not affected by temperature changes, making it more reliable for boiling point elevation and freezing point depression calculations.

When we were solving our exercise, molality was a stepping stone to finding the mass of solute. We calculated it by isolating \(m\) from the boiling point elevation equation: \(m = \frac{\Delta{T_b}}{K_b}\). Understanding this concept is crucial because it connects the amount of solute to the physical property of the solvent (in this case, boiling point elevation), illustrating its fundamental role in colligative properties.

Clausius-Clapeyron Equation

The Clausius-Clapeyron equation is instrumental in understanding phase changes, especially when dealing with boiling point and vapor pressure. This equation provides a way to quantify the relationship between the vapor pressure of a substance and its temperature. Here's the equation in its general form: \[\text{ln}(P_2/P_1) = -\frac{\text{Δ}H_{vap}}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right)\]

In the exercise, we simplified it using the assumption that the vapor pressure at the boiling point is equal to the surrounding atmospheric pressure, which permitted us to ignore the pressure terms and focus on solving for the normal boiling point of the solvent. Understanding this relationship is vital for students, as it ties together thermodynamic principles with observable properties like boiling point, thus enriching their comprehension of physical chemistry concepts.