Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogeneous solution. These are called colligative properties. Applications of colligative properties are very useful in day-today life. One of its examples is the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator of automobiles. A solution M is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is 0.9. Given: Freezing point depression constant of water ( K f text water )=1.86 K kg mol -1 Freezing point depression constant of ethanol ( K f text ethanol )=2.0 K kg mol -1 Boiling point elevation constant of water ( K b textwater )=0.52 K kg mol -1 Boiling point elevation constant of ethanol ( K b text ethanol )=1.2 K kg mol -1 Standard freezing point of water =273 K Standard freezing point of ethanol =155.7 K Standard boiling point of water =373 K Standard boiling point of ethanol =351.5 K Vapour pressure of pure water =32.8 mm Hg Vapour pressure of pure ethanol =40 mm Hg Molecular weight of water =18 g mol -1 Molecular weight of ethanol =46 g mol -1 In answering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non-volatile and nondissociative. Water is added to the solution M such that the fraction of water in the solution becomes 0.9. The boiling point of this solution is

Question

Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogeneous solution. These are called colligative properties. Applications of colligative properties are very useful in day-today life. One of its examples is the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator of automobiles. A solution M is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is 0.9. Given: Freezing point depression constant of water ( K f text water )=1.86 K kg mol -1 Freezing point depression constant of ethanol ( K f text ethanol )=2.0 K kg mol -1 Boiling point elevation constant of water ( K b textwater )=0.52 K kg mol -1 Boiling point elevation constant of ethanol ( K b text ethanol )=1.2 K kg mol -1 Standard freezing point of water =273 K Standard freezing point of ethanol =155.7 K Standard boiling point of water =373 K Standard boiling point of ethanol =351.5 K Vapour pressure of pure water =32.8 mm Hg Vapour pressure of pure ethanol =40 mm Hg Molecular weight of water =18 g mol -1 Molecular weight of ethanol =46 g mol -1 In answering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non-volatile and nondissociative. Water is added to the solution M such that the fraction of water in the solution becomes 0.9. The boiling point of this solution is (A) 380.4 K (B) 376.2 K (C) 375.5 K (D) 354.7 K

Tardigrade Admin · Accepted Answer

Δ T b = K b × m =0.52 × (0.1/0.9 × 18) × 1000=3.2 K T b =373+3.2=376.2 K

380.4 K

376.2 K

375.5 K

354.7 K

ΔTb​=Kb​×m =0.52×0.9×180.1​×1000=3.2K Tb​=373+3.2=376.2K

$Δ T_{b} = K_{b} \times m$
$= 0.52 \times \frac{0.1}{0.9 \times 18} \times 1000 = 3.2 K$
$T_{b} = 373 + 3.2 = 376.2 K$