Question

A solution of ethanol CH3CH2OH and water that is 10.% ethanol by mass is boiling at 98.7°C. The vapor is collected and cooled until it condenses to form a new solution.

Calculate the percent by mass of ethanol in the new solution. Here's some data you may need:

	normal boiling point	density	vapor pressure at 98.7°C
ethanol	78.°C	0.79gmL	1563.torr
water	100.°C	1.00gmL	725.torr

Be sure your answer has the correct number of significant digits.

Note for advanced students: you may assume the solution and vapor above it are ideal.

Answer 1

Sol. As solution is 10 % ethanol by mass

Mass of ethanol = 10 g

Mass of water = 90 g

Molar Mass of ethanol = 46.07 g/mol

Molar Mass of water = 18.01 g/mol  

So , Moles of ethanol = n1 = 10 g / 46.07 g/mol

= 0.2170 mol

Moles of water = n2 = 90 g / 18.01 g/mol

= 4.9972 mol

As Mole fraction of ethanol

= x1 = n1 / ( n1 + n2 )  

= 0.2170 mol / ( 0.2170 mol + 4.9972 mol )

= 0.2170 mol / 5.2142 mol

= 0.0416

Mole fraction of water = x2 = 1 - x1

= 1 - 0.0416 = 0.9584  

Now , Partial Pressure of pure ethanol = P1° = 1563 torr

Partial Pressure of pure water = P2° = 725 torr

So ,

Total Partial Pressure of solution

= P = P1° × x1 + P2° × x2

= 1563 torr × 0.0416 + 725 torr × 0.9584

= 759.8608 torr

Therefore , Mole fraction of ethanol in vapour phase

= y1 = P1° × x1 / P

= 1563 torr × 0.0416 / 759.8608 torr

= 0.0855

and , Mole fraction of water in vapour phase

= 1 - y1 = 1 - 0.0855 = 0.9145

Assuming the total number of moles are 1 in vapour phase

Moles of ethanol = 0.0855 × 1 = 0.0855 mol

Moles of water = 0.9145 × 1 = 0.9145 mol

So , Mass of ethanol in vapour phase

= 0.0855 mol × 46.07 g/mol = 3.9389 g

and , Mass of water in vapour phase

= 0.9145 mol × 18.01 g/mol = 16.4701 g

Therefore , Percent by mass of ethanol in the new solution

= ( 3.9389 g / ( 3.9389 g + 16.4701 g ) ) × 100

= ( 3.9389 g / 20.409 g ) × 100

=    19.30 %