Text Solution Verified by Experts The correct Answer is B ΔT f(observed) = 0.45∘ ΔT f(calculated) = 1000Kfw2 w1 ×M 2 = 1000× 5.12×0.2 20×60 (M 2(CH3COOH) = 60) = 0.853∘ ∴ = ΔT f(observed) ΔT f(calculated) = 0.45 0.853 = 0.527 2CH3COOH 1−α ⇔ (CH3COOH)2 α 2 Total moles = 1− α+ α 2 = 1 − α 2 i = 1− α 2 = 0.527 or α 2 = 0.473 or α = 0.946

Solution Freezing point: The freezing point of a substance may be defined as the temperature at which the vapour pressure of solid is equal to the vapour pressure of liquid. Concept: Colligative Properties and Determination of Molar Mass - Depression of Freezing Point Is there an error in this question or solution? Q 2.8 Q 2.7.4 Q 3.1

ΔT f = T f − T * f is the final freezing point (solution) minus the initial freezing point (solvent), and is always negative. ( * indicates pure solvent.) m is the molal concentration of the solution. ( molality = moles of solute / kg of solvent.) Kf is the freezing point depression constant and has a positive sign.

W 1. Because the internal energy E int of an ideal gas is a function of the temperature only, the change of the internal energy is zero, that is, Δ E int = 0 during this isothermal expansion. With the first law of thermodynamics, Δ E int = Q − W, we find that the heat absorbed by the gas is Q h = W 1 = n R T h ln V N V M.