18. what is the boiling point of a solution containing 0.80 g caffeine, c₈h₁₀n₄o₂, dissolved in 13.20 g…

Explanation:

Step1: Calculate moles of caffeine

Moles = mass / molar mass. Mass of caffeine is 0.80 g, molar mass is 194.0 g/mol. So moles = $\frac{0.80\ g}{194.0\ g/mol} \approx 0.004124\ mol$.

Step2: Calculate molality (m)

Molality = moles of solute / kg of solvent. Solvent (benzene) mass is 13.20 g = 0.01320 kg. So m = $\frac{0.004124\ mol}{0.01320\ kg} \approx 0.3124\ m$.

Step3: Calculate boiling point elevation ($\Delta T_b$)

$\Delta T_b = K_b \times m$. $K_b = 2.53\ ^\circ C/m$, m = 0.3124 m. So $\Delta T_b = 2.53\ ^\circ C/m \times 0.3124\ m \approx 0.790\ ^\circ C$.

Step4: Calculate boiling point of solution

Boiling point = boiling point of pure benzene + $\Delta T_b$. Pure benzene boiling point is $80.1\ ^\circ C$. So boiling point = $80.1\ ^\circ C + 0.790\ ^\circ C \approx 80.9\ ^\circ C$ (close to option C, considering rounding in steps). Wait, wait, recalculating step 2: 0.80 / 194 = 0.0041237. 0.0041237 / 0.0132 = 0.3124 m. Step 3: 2.53 * 0.3124 = 0.790. 80.1 + 0.790 = 80.89 ≈ 80.9? Wait, but let's check again. Wait, maybe I made a mistake. Wait, 0.80 g caffeine: moles = 0.80 / 194 = 0.0041237 mol. Solvent mass: 13.20 g = 0.0132 kg. Molality = 0.0041237 / 0.0132 = 0.3124 m. $\Delta T_b = 2.53 * 0.3124 = 0.790$. 80.1 + 0.790 = 80.89 ≈ 80.9, which is option C. But wait, maybe the calculation is more precise. Wait, 0.80 / 194 = 0.00412371134 mol. 0.00412371134 / 0.0132 = 0.312402374 m. 2.53 * 0.312402374 = 0.790378006. 80.1 + 0.790378006 = 80.890378 ≈ 80.9, which is option C.

Answer:

C. $80.9\ ^\circ C$