The density of aqueous solution of ethanol with mass percent of 33.4% is 0.951 g/mL. Calculate the molarity of c2h5oh in the solution. The molar mass of ethanol is 47 g/mol
Answers
Answer:We use a
1
⋅
m
L
volume of alcohol solution..........
We want
Moles of ethyl alcohol
Volume of solution
, we proceed on the basis of a
1
⋅
m
L
volume.........
1
a
.
1
⋅
m
L
×
0.984
⋅
g
⋅
m
L
−
1
×
10
%
46.07
⋅
g
⋅
m
o
l
−
1
1
⋅
m
L
×
1
×
10
−
3
⋅
L
⋅
m
L
−
1
=
2.14
⋅
m
o
l
⋅
L
−
1
1
b
.
We require
solution molality
=
moles of ethyl alcohol solute
kilograms of solvent
We already have the moles of solute in a
1
⋅
m
L
volume, and the mass of solvent is something we can determine with
ρ
.
1
⋅
m
L
×
0.984
⋅
g
⋅
m
L
−
1
×
10
%
46.07
⋅
g
⋅
m
o
l
−
1
90
%
×
0.984
⋅
g
⋅
m
L
−
1
×
1
⋅
m
L
×
10
−
3
⋅
k
g
⋅
g
−
1
=
2.42
⋅
m
o
l
⋅
k
g
−
1
.
c
.
We want
0.125
⋅
m
o
l
of ethyl alcohol, but solution concentration is
2.14
⋅
m
o
l
⋅
L
−
1
with respect to ethyl alcohol, and so we takes the quotient.....
0.125
⋅
m
o
l
2.14
⋅
m
o
l
⋅
L
−
1
=
0.0584
⋅
1
L
−
1
=
0.0584
⋅
L
×
10
3
⋅
m
L
⋅
L
−
1
=
58.4
⋅
m
L
d
.
We want mole fraction with respect to water......
χ
H
2
O
=
moles of water
moles of water + moles of ethanol
We can work, again, from a
1
⋅
m
L
volume..........
χ
H
2
O
=
0.984
⋅
g
×
0.9
18.01
⋅
g
⋅
m
o
l
−
1
0.984
⋅
g
×
0.9
18.01
⋅
g
⋅
m
o
l
−
1
+
0.984
⋅
g
×
0.1
46.07
⋅
g
⋅
m
o
l
−
1
=
0.958
And since, by definition,
χ
H
2
O
+
χ
ethanol
=
1
χ
ethanol
=
1
−
χ
H
2
O
=
1
−
0.958
=
0.042
.
You will have to check this calculation........
all care taken, but no responsibility admitted..
Explanation: