Start with what they give you and work towards what they are asking for, one
piece at a time.
We know that the atmospherics pressure is 760 mm Hg
source https://www.google.co.in/search?q=molar+mass+ocrtane&oq=molar+mass+ocrtan
e&aqs=chrome..69i57j0l2.3369j0j7&sourceid=chrome&espv=210&es_sm=122&i
e=UTF-8#q=1%20atm%20%3D%20mmhg&safe=off
The molar mass of octane is 114.23 gm/mol
source https://www.google.co.in/search?q=molar+mass+ocrtane&oq=molar+mass+ocrtan
e&aqs=chrome..69i57j0l2.3369j0j7&sourceid=chrome&espv=210&es_sm=122&i
e=UTF-8#q=molar+mass+octane&safe=off&spell=1
Molar mass of water is 18 gm/mol
source https://www.google.co.in/search?q=Molar+mass+of+water+is&oq=Molar+mass+o
f+water+is&aqs=chrome..69i57.138j0j7&sourceid=chrome&espv=210&es_sm=12
2&ie=UTF-8
Now
18 gm =1 mole
1gm H2O = 0.055mol
Now, Vapor Pressure of water at 90 degree celcius is 525.8 mmHg
source –
http://intro.chem.okstate.edu/1515sp01/database/vpwater.html
Now we have the equation,
x = p1/p,
where,
x = mole fraction of any individual component mixture
p1= partial pressure of any individual component in a mixture = 525.8 mmHg
p = total pressure = 760 mm Hg
Therefore
x = 525.8/733 = 0.7173
x = mole fraction of H2O = 0.7173
Also
x = n1/(n1 +n2)
x = mole fraction of any individual component mixture = 0.7169
n1 = number of moles of first component = 0.055mol
n2 = number of moles of second component
So now we substitute the value of x= 0.7169 in the above equation
0.7173 = x = n1/(n1 +n2)
Now n1 as calculated above = 1gm H2O = 0.055mol
Substitute this n1 in the equation
0.7173 = 0.055/(0.055 +n2)
Solving this equation
We get n2 = 0.02167
Therefore mass of octane that co-distills with each gram of water
= number of moles * molar mass (given above )
= 0.02167 * 114.23
= 2.47 gm
PART ii)
In the above part we calculated the molar composition of the distillation
%H2O = 0.7173 = 71.73 %
% octane = 1-0.7173 = 0.2827 = 28.31 %
Therefore
In 1 mole of mixture, mass of H2O = molar composition * molar mass of H2O
= 0.7173 * 18
= 12.9 gm
In 1 mole of mixture, mass of octane = molar composition * molar mass of octane
= 0.2831 * 114.23
= 32.33 gm
Therefore mass of 1 mole of mixture = 12.9 + 32.33 = 45.2 gm
%H2O = 12.9/45.2
= 0.2853
= 28.53% by mass
%octane = 32.33/45.2
= 0.7152
= 71.52% by mass
Therefore Percent composition of the vapour that is produced during the steam
distillation is
%H2O = 28.53
%octane = 71.52