"13.136 A pharmaceutical preparation made with ethanol (C2H5OH) is contaminated with methanol (CH3OH). A sample of vapor above the liquid mixture contains a 97/1 mass ratio of C2H5OH to CH3OH. What is the mass ratio of these alcohols in the liquid mixture

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