Respuesta :
Answer:
Mass ratio of ethanol to methanol in the liquid mixture = 202:1
Explanation:
Starting with a basis of 98.0 g
Amount of ethanol vapour in mass = 97 g
Amount of methanol vapour In mass = 1 g
Number of moles = (mass)/(molar mass)
Molar mass of ethanol = 46.07 g/mol
Molar mass of methanol = 32.04 g/mol
Number of moles of ethanol in the vapour = 97/(46.07) = 2.105 moles
Number of moles of methanol in the vapour = 1/(32.04) = 0.0312 mole
But we do know that
(Number of moles in gaseous form) = (number of moles in liquid state) × (Partial pressure).
Partial Pressure of ethanol = 60.5 torr
Partial pressure of methanol = 126.0 torr
(2.105) = (Number of moles of ethanol in the liquid state) × 60.5
number of moles of ethanol in the liquid state = 2.105/60.5 = 0.0348 moles/torr
(0.0312) = (Number of moles of methanol in the liquid state) × 126
number of moles of methanol in the liquid state = 0.0312/126 = 0.000248 moles/torr
We then convert these new number of moles in liquid state into masses
Mass = (Number of moles) × (molar mass)
(Mass of ethanol in the liquid state) = 0.0348 × 46.07 = 1.603 g/torr
(Mass of methanol in the liquid state) = 0.000248 × 32.04 = 0.00795 g/torr
Mass ratio of ethanol to methanol = (1.603/0.00795) = 201.7: 1 ≈ 202:1
