Question

1. what is the boiling point of a solution containing 0.80 g caffeine, c₈h₁₀n₄o₂, dissolved in 13.20 g benzene ? the boiling point of pure benzene is 80.1 °c and the boiling point elevation constant, kᵦ, is 2.53 °c/m. (molar mass c₈h₁₀n₄o₂ = 194.0 g/mol) a. 79.3 °c b. 80.4 °c c. 80.9 °c d. 85.2 °c e. 88.2 °c

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 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 $\Delta T_b = 2.53 \times 0.3124 \approx 0.790\ ^\circ C$.

Step4: Calculate boiling point of solution ($T_b$)

$T_b = T_{b,pure} + \Delta T_b$. $T_{b,pure} = 80.1\ ^\circ C$, $\Delta T_b \approx 0.790\ ^\circ C$. So $T_b = 80.1 + 0.790 \approx 80.9\ ^\circ C$ (close to option C, considering rounding during steps). Wait, wait, let's recalculate step2 more accurately: 0.80 / 194.0 = 0.0041237 mol. 0.0041237 / 0.01320 = 0.3124 m. Step3: 2.53 0.3124 = 2.530.3 + 2.53*0.0124 = 0.759 + 0.031372 = 0.790372. Step4: 80.1 + 0.790372 = 80.890372 ≈ 80.9 °C.

Answer:

C. 80.9 °C