please answer please
class 11 chemistry
What volume of ethanol is required to
make its Semimolar aqueous solution
having 500ml volume if the specific
gravity of methanol is 0.52.
..... ..... .....
please answer
Answers
Explanation:
I'll show you two methods that you can use to solve this problem.
a
THE MORE TEDIOUS APPROACH
a
As you know, a solution's percent concentration by mass tells you the number of grams of solute present for every
100 g
of solution.
To make the calculations easier, pick a
100-g
sample of this solution.
Now, you know that the mass of this sample will be equal to the mass of the ethanol, the solute, and the mass of the water, the solvent.
m
solution
=
m
ethanol
+
m
water
In your case, you will have
m
ethanol
+
m
water
=
100 g
(
1
)
You also know that the mole fraction of ethanol, which is defined as the ratio between the number of moles of ethanol and the total number of moles present in the solution, is equal to
0.25
.
χ
ethanol
=
n
ethanol
n
ethanol
+
n
water
At this point, you must use the molar masses of ethanol and of water to express the mole ratio of ethanol in terms of
m
ethanol
and
m
water
.
M
M ethanol
=
46.07 g mol
−
1
M
M water
=
18.015 g mol
−
1
This means that you have
n
ethanol
=
m
ethanol
46.07 g mol
−
1
n
water
=
m
water
18.015 g mol
−
1
Therefore, the mole fraction of ethanol can be rewritten as--for the sake of simplicity, I won't add any units
χ
ethanol
=
m
ethanol
46.07
m
ethanol
46.07
+
m
water
18.015
which is equivalent to
18.015
⋅
m
ethanol
18.015
⋅
m
ethanol
+
46.07
⋅
m
water
=
0.25
(
2
)
Now all you have to do is to solve this system of two equations with two unknowns.
Use equation
(
1
)
to write
m
water
=
100
−
m
ethanol
Plug this into equation
(
2
)
to find
18.015
⋅
m
ethanol
=
0.25
⋅
18.015
⋅
m
ethanol
+
0.25
⋅
46.07
⋅
(
100
−
m
ethanol
)
This will get you
m
ethanol
⋅
(
18.015
−
0.25
⋅
18.015
+
0.25
⋅
46.07
)
=
0.25
⋅
46.07
⋅
100
which results in
m
ethanol
=
1151.75
25.02875
=
46.02
Since this represents the mass of ethanol present in
100 g
of solution, you can say that the percent concentration by mass of ethanol is
% ethanol by mass = 46%
−−−−−−−−−−−−−−−−−−−−−−
a
THE LESS TEDIOUS APPROACH
a
Alternatively, you can start by picking a sample of this solution that contains exactly
1
mole solute and of solvent.
This means that you have
n
ethanol
+
n
water
=
1 mole
Now, you can use the mole fraction of ethanol to say that the number of moles of ethanol present in this sample is equal to
χ
ethanol
=
n
ethanol
1 mole
⇒
n
ethanol
=
0.25
⋅
1 mole
=
0.25 moles
Consequently, you can say that this sample contains
0.75
moles of water.
Use the molar masses of the two compounds to convert the number of moles to grams.
0.25
moles ethanol
⋅
46.07 g
1
mole ethanol
=
11.52 g
0.75
moles water
⋅
18.015 g
1
mole water
=
13.51 g
The total mass of the solution will be
11.52 g + 13.51 g = 25.03 g
You can use the known composition of the sample to figure out how many grams of ethanol you'd ge for
100 g
of this solution
100
g solution
⋅
11.52 g ethanol
25.03
g solution
=
46.02 g ethanol
Once again, you have
% ethanol by mass = 46%
−−−−−−−−−−−−−−−−−−−−−−
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