(a)(i) Ishan's automobile radiator is filled with 1.0 kg of water. How many grams of ethylene glycol (molar mass = 62 g mol) must Ishan add to get the freezing point of the solution lowered to −2.8°C? (K for water = 1.86 K kg mol)
(ii) What type of deviation from Raoult's law is shown by an ethanol and acetone mixture? Give reason.
(ii) What type of deviation from Raoult's law is shown by an ethanol and acetone mixture? Give reason.
Answer: (a)(i) ≈ 93.3 g (ii) Positive deviation — acetone breaks ethanol's hydrogen bonds, weakening the net attractions.
Solution
(i) Required molality:
Moles needed in 1.0 kg water = 1.505 mol.
Mass of glycol:
(ii) Positive deviation. Acetone molecules come between ethanol molecules and break their hydrogen bonds; the ethanol–acetone attractions are weaker than the original ethanol–ethanol attractions, so both components escape more readily and the vapour pressure exceeds the Raoult's-law value.
Moles needed in 1.0 kg water = 1.505 mol.
Mass of glycol:
(ii) Positive deviation. Acetone molecules come between ethanol molecules and break their hydrogen bonds; the ethanol–acetone attractions are weaker than the original ethanol–ethanol attractions, so both components escape more readily and the vapour pressure exceeds the Raoult's-law value.
CBSE Class 12 Board · 2024 · Subjective