Question

When $0.55\mathrm{g}$ of pure benzoic acid $\left({\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}\right)$ is dissolved in $32.0\mathrm{g}$ of benzene $\left({\mathrm{C}}_6{\mathrm{H}}_6\right)$, the freezing point of the solution is ${0.36}^{\circ }\mathrm{C}$ lower than the freezing point value of ${5.5}^{\circ }\mathrm{C}$ for the pure solvent. (a) Calculate the molecular weight of benzoic acid in benzene. (b) Use the structure of the solute to account for the observed value:

Short answer

The molecular weight of benzoic acid in benzene can be calculated using the freezing point depression formula and the given information, with the expression: $$\text{molecular weight}=\frac{0.55\mathrm{g}\times k_f}{{0.36}^{\circ }\mathrm{C}\times 0.032\mathrm{kg}}$$ However, we cannot obtain a numerical value without the cryoscopic constant ($k_f$) for benzene. The observed value of freezing point depression is 0.36ºC lower than the pure solvent due to the polar structure of benzoic acid, which forms a combination of Van der Waals forces and hydrogen bonding with nonpolar benzene.

Step-by-step solution

Understand Freezing Point Depression Formula

The formula to calculate freezing point depression is given by: $$\mathrm{\Delta }{T}_f=kf\times m$$ Where $\mathrm{\Delta }{T}_f$ = change in freezing point, $k_f$ = cryoscopic constant of the solvent, $m$ = molality of the solution. We also know: $$m=\frac{\text{moles of solute}}{\text{kilograms of solvent}}$$

Calculate the molality of the solution

First, we need to calculate the molality of the solution. We are given the mass of solute, mass of solvent, and the change in freezing point. Given: $\mathrm{\Delta }{T}_f={0.36}^{\circ }\mathrm{C}$, Mass of benzoic acid = $0.55\mathrm{g}$, Mass of benzene = $32.0\mathrm{g}$. First, let's find the moles of benzoic acid by dividing its mass by its molecular weight: $$\text{moles of benzoic acid}=\frac{0.55\mathrm{g}}{\text{molecular weight}}$$ Now, let's find the molality of the solution: $$m=\frac{\text{moles of benzoic acid}}{0.032\mathrm{kg}}$$

Calculate the cryoscopic constant of benzene

We are given the freezing point of pure benzene and the change in freezing point of benzene when the solute is added. The freezing point depression can be expressed as: $$\mathrm{\Delta }{T}_f=k_f\times m$$ Now, we need to find the cryoscopic constant ($k_f$) for benzene. The freezing point of pure benzene ($T_f^0$) is given as ${5.5}^{\circ }\mathrm{C}$. But we are not given the $k_f$ value, so we can't directly calculate it. Given the change in freezing point, we can at least express the freezing point depression as: $$\mathrm{\Delta }{T}_f=T_f^0-T_f={0.36}^{\circ }\mathrm{C}$$

Calculate the molecular weight of benzoic acid

Now that we have expressed freezing point depression, we can solve for the molecular weight of the benzoic acid. From step 2, we found that: $$m=\frac{\text{moles of benzoic acid}}{0.032\mathrm{kg}}$$ which can also be written as: $$m=\frac{0.55\mathrm{g}}{\text{molecular weight}\times 0.032\mathrm{kg}}$$ From step 3, we have: $$\mathrm{\Delta }{T}_f=k_f\times m$$ Combining these equations, we get: $${0.36}^{\circ }\mathrm{C}=k_f\times \frac{0.55\mathrm{g}}{\text{molecular weight}\times 0.032\mathrm{kg}}$$ We don't have the $k_f$ value, but we can still solve for the molecular weight by rearranging the equation: $$\text{molecular weight}=\frac{0.55\mathrm{g}\times k_f}{{0.36}^{\circ }\mathrm{C}\times 0.032\mathrm{kg}}$$ While we can't obtain a numerical value for the molecular weight of benzoic acid without the value of $k_f$, the expression demonstrates how the molecular weight can be calculated given the necessary constants. The answer for part (a) would be the expression above.

Analyze the structure of the solute

For part (b), we are asked to explain the observed value based on the structure of the solute. Benzoic acid is composed of a benzene ring attached to a carboxylic acid group (${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}$), which makes it polar and able to form hydrogen bonds. The solvation of benzoic acid in benzene results from a combination of Van der Waals forces and hydrogen bonding, given that benzene is nonpolar. Due to both types of interaction, the observed value of freezing point depression is 0.36ºC lower than the pure solvent, which is reasonable considering benzoic acid's molecular structure.