2.7 kg of each of water and acetic acid are mixed. The freezing point of the solution will be -x° C. Consider the acetic acid does not dimerise in water, nor dissociates in water. x= ________ (nearest integer)
[Given: Molar mass of water =18 g mol⁻¹, acetic acid =60 g mol⁻¹
^(K_f) H₂ O: 1.86 K kg mol⁻¹
K_f acetic acid: 3.90 K kg mol⁻¹
freezing point: H₂ O=273 K, acetic acid =290 K]

Solution

Chemistry ma’am · 3 steps

Equal masses but not equal moles: water has more moles, so water is the solvent here.

1. Count the moles. Water: 2700/18 = 150 mol. Acetic acid: 2700/60 = 45 mol. The component with more moles is the solvent, so water is the solvent and acetic acid is the solute.
2. Find the molality. Molality of acetic acid = (45 mol)/(2.7 kg) = 16.67 mol kg⁻¹. Use K_f of water, because water is the solvent.

   Δ T_f = K_f × m = 1.86 × 45/2.7 = 31 K

3. So. The solution freezes 31 K below 0 °C, at -31 °C, so x = 31.

The common slip: 217. Taking acetic acid as the solvent gives Δ T_f = 3.90 × 150/2.7 = 217, not 31. Water has more moles, so water is the solvent.

The idea to keep. When the masses are equal, compare moles to decide the solvent. Then use that solvent's K_f and that solvent's mass in kg for the molality.

Answer from the NTA final answer key. Solution by PrepWiser.