The normal boiling point of ethanol, $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH},\( is \)78.4^{\circ} \mathrm{C}\(. When \)3.26 \mathrm{~g}$ of a soluble nonelectrolyte is dissolved in \(100.0 \mathrm{~g}\) of ethanol at that temperature, the vapor pressure of the solution is \(100 \mathrm{kPa}\). What is the molar mass of the solute?

According to Raoult's law, the vapor pressure of a solution equals the mole fraction of the solvent multiplied by the vapor pressure of the pure solvent. \[P = X_A P^*_A\] Where: \(P\) = vapor pressure of the solution \(P^*_A\) = vapor pressure of the pure solvent \(X_A\) = mole fraction of the solvent

First, we convert the vapor pressure of the solution to the same units as the boiling point of ethanol. \[\text{Vapor Pressure of the solution} = 100\, kPa = 760 \, mmHg \] Next, we will calculate the mole fraction of the solvent. Rearrange Raoult's law formula to find \(X_A\): \[X_A = \frac{P}{P^*_A}\] Now, we plug in the given values: \[X_A = \frac{760\,mmHg}{760\,mmHg} = 1\] So the mole fraction of the solvent is 1, which means that the mole fraction of the solute is 0.

Now we can calculate the molality (m) of the solution using the mass of solute dissolved in the solvent. \[m = \frac{moles \,of \,solute}{kg\,of\, solvent}\] We are given: - mass of solute: 3.26 g - mass of solvent (ethanol): 100.0 g = 0.100 kg To find the moles of solute, we need to find the molar mass of the solute. Let's represent the molar mass of solute by "M". \[moles \,of\,solute = \frac{3.26\,g}{M}\] Plug in the given values: \[m = \frac{\frac{3.26\,g}{M}}{0.100\,kg}\]

We know that the mole fraction of the solute is 0, so the molality of the solution can also be represented as: \[m = \frac{moles \,of \,solute}{moles \,of \,solvent}\] Now, we can use the mole fraction and molality of the solution to find the molar mass of solute (M): \(m = \frac{\frac{3.26\,g}{M}}{0.100\,kg} = \frac{moles \,of \,solute}{moles \,of \,solvent}\) Solving for M: \[M = \frac{3.26\,g \times moles \,of \,solvent}{0.100\,kg \times moles \,of \,solute}\] Now let's plug in the mole fraction of solute, which is 0: \[M = \frac{3.26\,g \times 0}{0.100\,kg \times 0}\] Since the mole fraction of solute is 0, we get an indeterminate form. However, we know that the mole fraction of solute cannot be exactly zero because there is a presence of solute in the solution. Thus, there might be an error in the given information or a small deviation in the mole fraction. Therefore, we cannot determine the molar mass of the solute with the given information.