Calculate the freezing point of a solution made from 22.0 g of octane (C₈H₁₈)dissolved in 148.0 g of benzene. Benzene freezes at 5.50°C and its K_f value is 5.12°C/m. A)-1.16°C B)0.98°C C)6.66°C D)12.2°C E)5.49°C

Question

Quizplus · Accepted Answer

First, we need to calculate the molality of the solution:

molality = moles of solute / kg of solvent

The molar mass of octane is 114.23 g/mol, so:

moles of octane = 22.0 g / 114.23 g/mol = 0.193 mol

kg of benzene = 148.0 g / 1000 g/kg = 0.148 kg

molality = 0.193 mol / 0.148 kg = 1.30 m

Next, we can calculate the change in freezing point:

ΔTf = Kf x molality

ΔTf = 5.12°C/m x 1.30 m = 6.656°C

Finally, we can use this value to calculate the freezing point of the solution:

freezing point = 5.50°C - 6.656°C = -1.156°C

Therefore, the best choice is A) -1.16°C.

Calculate the Freezing Point of a Solution Made from 22