Question

Eugenol, the active ingredient of cloves, has a formula of C10H12O2. What is the boiling point of a solution containing 0.144 g of this compound dissolved in 10.0 g. of benzene? The Kb for benzene is 2.53 oC/m, and its normal boiling point is 80.10 oC.

Answer 1

To calculate the boiling point of a 0.144 g eugenol solution in 10.0 g of benzene, first calculate the molality, then use the ebullioscopic constant to find the boiling point elevation. Adding this to benzene's normal boiling point gives a solution boiling point of approximately 80.10022 °C.

Step-by-step explanation:

The boiling point of a solution containing 0.144 g of eugenol (C10H12O2) dissolved in 10.0 g of benzene can be calculated using the formula for boiling point elevation ΔTb = iKbm, where ΔTb is the boiling point elevation, Kb is the ebullioscopic constant of benzene, m is the molality of the solution, and i is the van't Hoff factor (which is 1 for non-electrolytes like eugenol).

First, we calculate the molality (m) of the eugenol solution, which is moles of eugenol per kilogram of benzene. The molar mass of eugenol is 164.20 g/mol, so we have:

molality (m) = moles of eugenol / mass of benzene in kg
= (0.144 g eugenol / 164.20 g/mol) / (10.0 g benzene / 1000 g/kg)
= 8.77 x 10-5 mol/kg

Then, the boiling point elevation can be calculated:

ΔTb = (1)(2.53 °C/m)(8.77 x 10-5 mol/kg)
= 0.00022201 °C

Adding this elevation to the normal boiling point of benzene (80.10 °C), gives us the boiling point of the eugenol solution:

Boiling point of eugenol solution: 80.10 °C + 0.00022201 °C ≈ 80.10022 °C

Answer 2

80.32 °C

Step-by-step explanation:

The difference in the boiling point is

ΔT= Kb * m

m stands for molality, moles of solute (in this case eugenol) divided by kg of solvent (in this case benzene).

We're given Kb already, so in order to calculate ΔT we calculate m first:

• 10.0 g of benzene = 0.01 kg bencene
• The molecular weight of eugenol is 12*10 + 12*1 + 2*16 = 164 g/mol

• 0.144 g eugenol ÷ 164 g/mol = 8.78 *10⁻⁴ mol eugenol.

m = 8.78 *10⁻⁴ mol / 0.01 kg = 0.0878 m

Finally we calculate ΔT:

ΔT = 2.53 °C/m * 0.0878 m = 0.22 °C

Thus the boiling point of the solution is

T₂ = T₁ + ΔT

T₂ = 80.10°C + 0.22°C

T₂ = 80.32 °C