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CBSE Questions for Class 12 Engineering Chemistry Solutions Quiz 5 - MCQExams.com
CBSE
Class 12 Engineering Chemistry
Solutions
Quiz 5
At
25
o
C
, the total pressure of an ideal solution obtained by mixing 3 mole of A and 2 mole of B, is 184 torr. What is the vapour pressure (in torr) of pure B at the same temperature (Vapour pressure of pure A at
25
o
C
is 200 torr.)?
Report Question
0%
180
0%
160
0%
16
0%
100
Explanation
x
A
=
n
A
n
A
+
n
B
=
3
3
+
2
=
3
5
=
0.6
x
B
=
0.4
p
=
p
0
A
+
p
0
B
x
B
184
=
200
×
0.6
+
p
0
B
×
0.4
184
=
120
+
p
0
B
×
0.4
p
0
B
=
64
0.4
=
160
t
o
r
r
An aqueous solution containing
28
%
by mass of a liquid A (molecular mass = 140) has a vapour pressure of 160 mm at
37
o
C
. Find the vapour pressure of pure liquid A (the vapour pressure of water
37
o
C
is 150 mm).
Report Question
0%
p
A
=
358.3
m
m
0%
p
A
=
388.3
m
m
0%
p
A
=
258.3
m
m
0%
None of these
Among the following, that does not form an ideal solution is:
Report Question
0%
C
6
H
6
and
C
6
H
5
C
H
3
0%
C
2
H
5
C
l
and
C
2
H
5
O
H
0%
C
6
H
5
C
l
and
C
6
H
5
B
r
0%
C
2
H
5
B
r
and
C
2
H
5
I
The depressions in freezing point for 1 m urea, 1 m glucose and 1 m NaCl are in the ratio:
Report Question
0%
1 : 2 : 3
0%
3 : 2 : 2
0%
1 : 1 : 2
0%
None of these
Explanation
We have,
Δ
T
f
=
i
K
f
m
i
u
r
e
a
=
i
g
l
u
c
o
s
e
=
1
,
i
N
a
C
l
=
2
So, we have ratio for
Δ
T
f
-
1
×
K
f
×
1
:
1
×
K
f
×
1
:
2
×
K
f
×
1
=
1
:
1
:
2
Hence, option (C) is correct.
Which one of the following is a primary standard?
Report Question
0%
Oxalic acid
0%
Sodium thiosulphate
0%
Sodium hydroxide
0%
Potassium dichromate
Explanation
The primary standard is a compound of sufficient purity from which standard solutions of known normalities can be prepared by direct weighing of it and diluting to a defined volume of solution.
Potassium Dichromate
(
K
2
C
r
2
O
7
)
is suitable to be used as a primary standard. It cannot be obtained in very pure form. It readily reacts with any traces of organic material or any other reducing substance in water.
18 g glucose is dissolved in 90 g of water. The relative lowering of vapour pressure of the solution is equal to__________.
Report Question
0%
6
0%
0.2
0%
5.1
0%
0.02
Explanation
Molecular mass of water
=
2
×
1
+
1
×
16
=
18
g
=2×1+1×16=18g
For
90
g
178.2g
water
n
A
=
5
Molecular mass of glucose
6
×
12
+
12
×
1
+
6
×
16
=
180
g
For
18
g
glucose
n
B
=
0.1
Now,
X
B
=
0.1
0.1
+
5
=
0.0196
~
0.02
X is a non-volatile solute and Y is volatile solvent. The following vapour pressures are obtained by dissolving X in Y.
X/mol
L
−
1
Y/mm Hg
0.1
P
1
0.25
P
2
0.01
P
3
The correct order of vapour pressure is :
Report Question
0%
P
1
<
P
2
<
P
3
0%
P
3
<
P
2
<
P
1
0%
P
3
<
P
1
<
P
2
0%
P
2
<
P
1
<
P
3
Explanation
Lesser is the concentration of solution greater is the vapour pressure
P
2
<
P
1
<
P
3
(
0.25
M
)
(
0.1
M
)
(
0.01
M
)
Hence, option B is correct.
The freezing point depression constant for water is
1.86
o
C
m
−
1
. If
5
g
N
a
2
S
O
4
is dissolved in 45 g
H
2
O
, the freezing point is changed by
−
3.82
o
C
.
Calculate the van't Hoff factor for
N
a
2
S
O
4
.
Report Question
0%
0.381
0%
2.05
0%
2.63
0%
3.11
Explanation
Given:
K
f
=
1.86
o
C
m
−
1
,
w
B
=
5
g
,
w
A
=
45
g
,
M
B
=
142
g
/
m
o
l
△
T
=
i
×
K
f
×
w
B
×
1000
M
B
×
w
A
3.82
=
i
×
1.86
×
5
×
1000
142
×
45
i
=
2.63
An aqueous solution is
1
molal in
K
I
. Which change will cause the vapour pressure of the solution to increase?
Report Question
0%
Addition of
N
a
C
l
0%
Addition of
N
a
2
S
O
4
0%
Addition of
1
molal
K
I
0%
Addition of water
If the various terms in the following expressions have usual meanings, the van't Hoff factor 'i' cannot be calculated by which of the following expression?
Report Question
0%
π
V
=
√
i
n
R
T
0%
△
T
f
=
i
×
K
f
×
m
0%
△
T
b
=
i
×
K
b
×
m
0%
p
o
s
o
l
v
e
n
t
−
p
s
o
l
u
t
i
o
n
p
o
s
o
l
v
e
n
t
=
i
[
n
N
+
n
]
Which of the following property indicates weak intermolecular forces of attraction in liquid?
Report Question
0%
High heat of vaporization
0%
High vapour pressure
0%
High critical temperature
0%
High boiling point
Explanation
The molecules that are having weak intermolecular forces of attraction, they can easily dissociate and convert into vapour state,
so, molecules with weak intractions have high vapour pressure.
Hence, option
(
B
)
is correct.
Choose the correct statement, when the concentration of a salt solution is increased:
Report Question
0%
Boiling point increases while vapour pressure decreases
0%
Boiling point decreases while vapour pressure increases
0%
Freezing point decreases while vapour pressure increases
0%
Freezing point increases while vapour pressure decreases
Explanation
The
vapour
pressure depends upon the escape of solvent molecules from the surface of the liquid. When
concentration
of
salt
solution is increased,
lesser
number of solvent molecules will escape into
vapours
. It means,
vapour
pressure of the solution will be less than that of the pure solvent and the boiling point of the solution is inversely proportional to the
vapour
pressure.
Boiling point
∝
1
V
a
p
o
u
r
p
r
e
s
s
u
r
e
That is why volatile solvent having a very low B.P. possesses a high vapour pressure.
Two liquids
A
and
B
form an ideal solution. At
300
K
,
the VP of solution containing one mole of
′
A
′
and
4
mole
′
B
′
is
560
m
m
H
g
.
At the same temp, if one mole of
′
B
′
is taken out from the solution the VP of the solution has decreased by
10
m
m
H
g
,
the VP of pure
A
&
B
are (in mm):
Report Question
0%
400
,
600
0%
500
,
500
0%
300
,
700
0%
200
,
800
If in a solvent, n simple molecules of solute combine to form an associated molecule,
α
is the degree of association, the van't Hoff's factor is equal to :
Report Question
0%
1
1
−
n
α
0%
1
−
α
+
n
α
0%
1
−
α
+
α
n
0%
α
n
−
1
+
α
Explanation
n
A
⇌
A
n
1
−
α
α
n
i
=
1
−
α
+
α
n
Hence, option C is correct.
Van't Hoff factor, when benzoic acid is dissolved in benzene, will be:
Report Question
0%
2
0%
1
0%
0.5
0%
1.5
Explanation
Benzoic acid dissolved in benzene is found to have molecular mass double of its normal molecular mass. This is explained by the fact that benzoic acid forms a dimer in benzene solution due to hydrogen bonding.
i
=
n (observed)
n (theoretical)
i
=
1
2
i
=
0.5
n (observed)
=
number of solute particles present in solution.
n (theoretical)
=
number of solute particles without considering association/dissociation.
Van't Hoff factor, when benzoic acid is dissolved in benzene, will be
0.5
.
Hence, the correct answer is
option C
.
The vapour pressure of a solvent decreased by
10
m
m
of mercury when a non-volatile solute was added to the solvent. The mole fraction of the solute in the solution is
0.2
. What should be the mole fraction of the solvent, if the decrease in the vapour pressure is to be
20
m
m
of mercury?
Report Question
0%
0.8
0%
0.6
0%
0.4
0%
0.2
Explanation
Mole fraction of solute
=
lowering in vapour pressure
vapour pressure of solvent
comparing under the two conditions
0.2
mole fraction of solute
=
10
20
or, the mole fraction of solute
=
0.4
mole fraction of solvent
=
(
1
−
0.4
)
=
0.6
Hence, the correct option is
B
At a constant temperature, which of the following aqueous solutions will have the maximum vapour pressure?
[Mol weight:
N
a
C
l
=
58.5
,
H
2
S
O
4
=
98.0
g
.
m
o
l
−
1
]
Report Question
0%
1
molal
N
a
C
l
0%
1
molar
N
a
C
l
0%
1
molal
H
2
S
O
4
0%
1
molar
H
2
S
O
4
Explanation
One molar (1M) aqueous solution is more concentrated than one molal aqueous solution of the same solute.
In solution,
H
2
S
O
4
provides three ions, while
N
a
C
l
provides two ions. Hence, the
vapour
pressure of a
solution
of
N
a
C
l
is higher (as it gives
less number of
ions).
Therefore, 1 molal
N
a
C
l
will have the maximum
vapour
pressure.
Hence, option
A
is correct.
Two liquids
A
and
B
form an ideal solution. What is the vapour pressure of solution containing
2
moles of
A
and
3
moles of
B
at
300
K
?
[Given: At
300
K
, vapour pressure of pure liquid
A
(
P
∘
A
)
=
100
t
o
r
r
, Vapour pressure of pure liquid
B
(
P
∘
B
)
=
300
t
o
r
r
]
.
Report Question
0%
200
t
o
r
r
0%
220
t
o
r
r
0%
180
t
o
r
r
0%
None of these
Explanation
x
A
=
2
2
+
3
=
2
5
,
x
B
=
3
5
Vapour pressure of solution
=
x
A
P
o
A
+
x
B
P
o
B
[Raoult's Law]
=
2
5
×
100
+
3
5
×
300
=
220
t
o
r
r
A cylinder filled with a movable piston contains liquid water in equilibrium with water vapour at
25
o
C
. Which one of the following operations results in a decrease in the equilibrium vapour pressure?
Report Question
0%
Moving piston downward a short distance
0%
Removing a small amount of vapour
0%
Removing a small amount of the liquid water
0%
Dissolving salt in the water
Explanation
Vapour pressure is a surface phenomenon, when the surface area is more, vapour pressure will be more. When we add solute, what we do is we decrease the vapour pressure of solvent because same of the salt ions will be present at the solvent surface, thus taking place of same of the solvent particles which were actively participating in maintain, that previous high value of vapour pressure. So, the equilibrium vapour pressure value decreases. Thus, dissolving salt in water results in decreasing of equilibrium vapour pressure.
On the surface of the earth at
1
a
t
m
pressure, a balloon filled with
H
2
gas occupies
500
m
L
. This volume is
5
/
6
of its maximum capacity. The balloon is left in air. It starts rising. The height above which the balloon will burst if temperature of the atmosphere remains constant and the pressure decreases
1
m
m
for every
100
c
m
rise of height is:
Report Question
0%
120
m
0%
136.67
m
0%
126
m
0%
100
m
Explanation
Given:
P
1
=
1
a
t
m
=
760
m
m
o
f
H
g
V
1
=
500
m
l
Solution:
Let Maximum capacity
V
2
= x
According to question ,
500
=
5
6
×
x
⇒
x
=
500
×
6
5
⇒
x
=
600
From the ideal gas law ,
P
1
V
1
=
P
2
V
2
⇒
760
×
500
=
P
2
×
600
⇒
P
2
=
760
×
500
600
⇒
P
2
=
633.33
m
m
o
f
H
g
Change in pressure =
P
2
−
P
1
⇒
C
h
a
n
g
e
i
n
p
r
e
s
s
u
r
e
=
760
−
633.33
⇒
C
h
a
n
g
e
i
n
p
r
e
s
s
u
r
e
=
126.67
m
m
o
f
H
g
It is given that for each 1 mm decrease, rise = 100 cm
Therefore, Height = 100 × 126.67 cm
⇒
Height = 126.67 m
Hence, option C is correct and the answer is 126.67m.
Lowering of vapour pressure in
2
molal aqueous solution at
373
K
is____________.
Report Question
0%
0.35
bar
0%
0.022
bar
0%
0.22
bar
0%
0.035
bar
Explanation
Given
A solute B have 2 moles in 1 kg of water
Solution
Molality=1
No.of moles of B=Xb
Molality=Xb*1000/(1-Xb)*18
Putting values
Xb=0.0347
No.of moles of Water=Xa
Xa=1-Xb
Xa=0.9653
Pressure of 1 mole =760mmHg
Pressure of 0.9653 mol=760*0.9653=733.628mmHg
Lowering of Vapour pressure=760-733.628=26.372mmHg=0.035 bar
The correct option is D
The vapour pressure of pure liquid
A
at
300
K
is
577
T
o
r
r
and that of pure liquid
B
is
390
T
o
r
r
. These two compounds form ideal liquid and gaseous mixtures. Consider the equilibrium composition of a mixture in which the mole fraction of
A
in the vapour is
0.35
. Find the mole % of
A
in liquid.
Report Question
0%
0.628
0%
0.872
0%
0.267
0%
0.834
Explanation
P
o
A
=
577
t
o
r
r
,
P
o
B
=
390
t
o
r
r
,
x
A
′
=
0.35
Mole fraction of
A
in vapour phase=
P
A
P
P
A
=
x
A
P
o
A
,
P
=
P
A
+
P
B
=
x
A
P
o
A
+
(
1
−
x
A
)
P
o
B
∴
0.35
=
x
A
×
577
x
A
×
577
+
(
1
−
x
A
)
390
∴
x
A
=
0.267
The freezing point depression constant for water is
−
1.86
C
. If
5.00
g of
N
a
2
S
O
4
is dissolved in
45.0
g
H
2
O
. The freezing point is changed by
−
3.82
C
.
Calculate the van't Hoff factor?
Report Question
0%
2.63
0%
0.311
0%
0.38
0%
3.8
Explanation
Depression in freezing point=
Δ
T
f
=
i
K
f
m
Given values,
Δ
T
f
=
−
3.82
o
C
K
f
=
−
1.86
o
C
=
?
Molality=
5
142
×
1000
45
=
0.7824
m
Δ
T
f
=
i
K
f
m
⇒
i
=
Δ
T
f
K
f
m
=
−
3.82
−
1.86
×
0.7824
⇒
i
=
3.82
1.455
=
2.62
For the reversible reaction.
N
2
(
g
)
+
3
H
2
(
g
)
⇌
2
N
H
3
(
g
)
at
500
o
C, the value of
K
p
is
1.44
×
10
−
5
a
t
m
−
2
when partial pressure is measured in atmosphere. The corresponding value of
K
c
, with concentration in
m
o
l
/
L
is:
Report Question
0%
1.44
×
10
−
5
(
0.082
×
500
)
−
2
0%
1.44
×
10
−
5
(
8.314
×
773
)
−
2
0%
1.44
×
10
−
5
(
0.082
×
773
)
2
0%
1.44
×
10
−
5
(
0.082
×
773
)
−
2
Explanation
For any given equilibrium,
K
P
=
K
C
(
R
T
)
Δ
n
g
where,
Δ
n
g
= (gaseous moles of products)-(gaseous moles of reactants)
In given reaction,
Δ
n
g
=
2
−
(
3
+
1
)
=
−
2
∴
K
P
=
K
C
(
R
T
)
2
∴
K
C
=
K
P
(
R
T
)
2
=
K
P
(
R
T
)
−
2
=
1.44
×
10
−
5
(
0.082
×
773
)
−
2
m
o
l
/
L
If relative decrease in vapour pressure is
0.4
for a solution containing
1
mol
N
a
C
l
in
3
mol
H
2
O
.
N
a
C
l
is_____ % ionized.
Report Question
0%
60
%
0%
50
%
0%
100
%
0%
40
%
Explanation
no. of moles of
N
a
C
l
=1
relative lowering of vapour pressure =
i
×
mole fraction of solute
relative lowering of vapour pressure=
i
×
n
(
s
o
l
u
t
e
)
n
(
s
o
l
u
t
e
)
+
n
(
s
o
l
v
e
n
t
)
0.4=
i
×
1
4
therefore
i
=
1.6
degree of dissociation/ionisation=
i
−
1
n
−
1
N
a
C
l
→
N
a
+
+
C
l
−
So,
n
=
2
α
=
0.6
=
60
%
Option A is correct.
Calculate solubility of
P
b
I
2
(
K
s
p
=
1
/
4
×
10
−
8
) in water at
25
o
, which is
90
% dissociated.
Report Question
0%
7.5
×
10
−
4
M
0%
1.6
×
10
−
3
M
0%
3.4
×
10
−
3
M
0%
8.2
×
10
−
3
M
Water and chlorobenzene are immiscible liquids. Their mixture boils at
89
o
C under a reduced pressure of
7.7
×
10
4
Pa. The vapour pressure of pure water at
89
o
C is
7
×
10
4
Pa. Weight percent of chlorobenzene in the distillate is:
Report Question
0%
50
0%
60
0%
79
0%
38.46
Explanation
Vapour pressure =
7.7
×
10
4
Pa
Total pressure =
7
×
10
4
Pa
Partial pressure = Vapour pressure = Mole fraction x Total pressure
Mole fraction of
H
2
O
=
V
a
p
o
u
r
p
r
e
s
s
u
r
e
T
o
t
a
l
p
r
e
s
s
u
r
e
=
7.7
×
10
4
7
×
10
4
= 0.909
Mole fraction of
C
H
C
l
3
= 1 - 0.909 = 0.091
Molar mass of
H
2
O
= 18g/mol
Molar mass of
C
H
C
l
3
= 119.5g/mol
Weight percent of chloroform =
Mole fraction of
C
H
C
l
3
x
Molar mass of
C
H
C
l
3
/ [
Molar mass of
C
H
C
l
3
+
Molar mass of
H
2
O
}
Weight percent of chloroform = 79%
Answer is 79%.
The
K
s
p
for
C
r
(
O
H
)
3
is
1.6
×
10
−
30
. The solubility of this compound in water is:
Report Question
0%
4
√
1.6
×
10
−
30
0%
4
√
1.6
×
10
−
30
/
27
0%
1.6
×
10
−
9
0%
2
√
1.6
×
10
−
30
Explanation
Let S M be the solubility of
C
r
(
O
H
)
3
[
C
r
3
+
]
=
[
C
r
(
O
H
)
3
]
=
S
[
O
H
−
]
=
3
[
C
r
(
O
H
)
3
]
=
3
S
The solubility product
K
s
p
=
[
C
r
3
+
]
[
O
H
−
]
3
1.6
×
10
−
30
=
S
×
(
3
S
)
3
1.6
×
10
−
30
=
27
S
4
S
=
[
1.6
×
10
−
30
27
]
1
/
4
S
=
3.6
×
10
−
8
M
Hence, solubility of
C
r
(
O
H
)
3
in water is
3.6
×
10
−
8
M
The vapour pressure of the solution of two liquids
A
(
p
o
=
80
m
m
) and
B
(
p
o
=
120
m
m
)
is found to be
100
m
m
when
x
A
=
0.4
. The result shows that:
Report Question
0%
solution exhibits ideal behaviour
0%
solution shows positive deviations
0%
solution shows negative deviations
0%
solution will show positive deviations for lower concentration and negative deviations for higher concentrations
Explanation
P
0
A
=
80
m
m
P
∘
B
=
120
m
m
x
n
=
0.4
x
B
=
0.6
P
T
=
x
A
P
0
A
+
x
B
P
0
0
=
0.4
×
80
+
0.6
×
120
=
32
+
72
P
T
=
104
Since the given total pressure is less than the calculated Total pressure therefore, it shows a negative deviation hence option C is correct.
A
and
B
form ideal solutions, at
50
o
C
,
P
o
A
is half
P
o
B
. A solution containing
0.2
mole of
A
and
0.8
mole of
B
has a normal boiling point of
50
o
C
. Find
18
×
P
o
B
. (
P
o
B
is in atm)
Report Question
0%
0.24
0%
0.34
0%
0.53
0%
0.46
The ratio of the vapour pressure of a solution to the vapour pressure of the solvent is:
Report Question
0%
equal to the mole fraction of the solvent
0%
equal to the mole fraction of the solute
0%
directly proportional to mole fraction of the solute
0%
None of the above
Vapour Pressure of a mixture of benzene and toluene is given by
P
=
179
X
B
+
92
, Where
X
B
is mole fraction of benzene.
If Vapours are removed and condensed in to liquid then what will be the ratio of mole fraction of benzene and toluene in first condensate :
Report Question
0%
2.8
0%
1.5
0%
3.5
0%
4.5
Explanation
Given
P
=
179
X
B
+
92
For pure
C
6
H
6
,
X
B
=
1
∴
P
o
B
=
179
+
92
=
271
m
m
For pure
C
7
H
8
,
X
B
=
0
∴
P
o
T
=
179
×
0
+
92
=
92
m
m
Now,
P
M
=
P
o
B
X
B
+
P
o
T
X
T
=
(
271
×
12
12
+
8
)
+
(
92
×
8
12
+
8
)
=
199.4
m
m
as, Moles of
C
6
H
6
=
936
78
=
12
Moles of
C
7
H
8
=
736
92
=
8
Now, mole fraction of
C
6
H
6
in vapour phase of initial mixture
X
1
B
=
162.6
199.4
and that of toulene,
X
1
T
=
36.8
199.4
∴
X
1
B
X
1
T
=
162.6
36.8
=
4.418
What is the solubility of
A
l
(
O
H
3
)
,
K
s
p
=
1
×
10
−
33
, in a solution having
p
H
=
4
?
Report Question
0%
6
×
10
−
3
M
0%
10
−
6
M
0%
1.5
×
10
−
4
M
0%
2.47
×
10
−
9
M
Explanation
The solubility equilibrium of
A
l
(
O
H
)
3
is
A
l
(
O
H
)
3
→
←
A
l
+
3
+
3
O
H
−
Therefore
x
=
1
,
y
=
2
and
K
s
p
=
[
A
l
+
3
]
×
[
O
H
−
]
=
27
S
4
S
=
(
10
−
33
27
)
1
/
4
Therefore
S
=
2.47
×
10
−
9
M
in
m
o
l
d
m
−
3
At a given temperature, total vapour pressure in Torr of a mixture volatile components A and B is given by
P
T
o
t
a
l
=
120
−
75
X
B
Hence, vapour pressure of pure A and B respectively (in Torr) are ?
Report Question
0%
120, 75
0%
120, 195
0%
120, 45
0%
75, 45
Explanation
P
T
o
t
a
l
=
P
A
X
A
+
P
B
X
B
But
X
A
=
1
−
X
B
(
∵
X
A
+
X
B
=
1
)
Hence,
P
T
o
t
a
l
=
P
A
(
1
−
X
B
)
+
P
B
X
B
P
T
o
t
a
l
=
P
A
−
P
A
X
B
+
P
B
X
B
P
T
o
t
a
l
=
P
A
−
(
P
A
−
P
B
)
X
B
But,
P
T
o
t
a
l
=
120
−
75
X
B
Hence,
P
A
=
120
t
o
r
r
P
A
−
P
B
=
75
P
B
=
P
A
−
75
P
B
=
120
−
75
P
B
=
45
t
o
r
r
Vapour Pressure of a mixture of benzene and toluene is given by
P
=
179
X
B
+
92
, Where
X
B
is mole fraction of benzene.
Vapour pressure of the solution obtained by mixing 936 g of benzene and 736 gm of toluene :
Report Question
0%
199.4 mm
0%
271 mm
0%
280 mm
0%
289 mm
Explanation
P=179
X
B
+92
If [Benzene] =936g
If[Toluene]=736g
X
B
= mole fraction of benzene in liquid
X
B
=
936
78
936
78
+
736
92
=0.6
Now
P
B
=179
×
0.6
+92
P
B
=199.4
At
80
0
C, the vapour pressure of pure liquid A is
250
mm of Hg and that of pure liquid B is
1000
mm of Hg. If a solution of A and B boils at
80
0
C and
1
atm pressure, the amount of A in the mixture is :
(
1
a
t
m
=
760
m
m
Hg)
Report Question
0%
50
mole percent
0%
52
mole percent
0%
32
mole percent
0%
48
mole percent
Explanation
1 atm = 760 mm Hg =
P
T
P
T
=
P
0
A
X
A
×
P
0
B
X
B
760
=
250
X
A
+
1000
(
1
−
X
A
)
240
=
750
X
A
X
A
=
0.32
Hence mole
%
of A =
32
%
The van't Hoff factor of the benzoic acid solution in benzene is 0.In this solution, benzoic acid:
Report Question
0%
dissociates
0%
forms dimer
0%
remains unchanged
0%
forms tetramer
Explanation
As colligative property
∝
n
(
no. of moles
)
Therefore,
i
(
van't Hoff factor
)
=
Observed number of particles
Theoretical number of particles
But
i
=
1
2
, therefore, Observed number of particles <
Theoretical number of particles
Thus, the benzoic acid forms a dimer.
The vapour pressure of the solution of two liquids
A
(
P
∘
=
80
m
m
)
and
B
(
p
∘
=
120
m
m
)
is found to be
100
m
m
when
x
A
=
0.4
. The result shows that :
Report Question
0%
Solution exhibits ideal behaviour
0%
Solution shows positive deviations
0%
Solution shows negative deviations
0%
Solution will show positive deviations for lower concentration and negative deviations for higher concentrations
Explanation
P
o
A
=
80
m
m
P
o
B
=
120
m
m
x
A
=
0.4
x
B
=
0.6
P
=
80
×
0.4
+
120
×
0.6
=
104
m
m
observed vapour pressure < calculated vapour pressure
(solution shows negative deviation
At
80
o
C, the vapour pressure of pure liquid
A
is
250
mm of
H
g
and that of pure liquid
B
is
1000
mm of
H
g
. If a solution of
A
and
B
boils at
80
o
and
1
atm pressure, the amount of
A
in the mixture is?
(
1
atm
=
760
mm Hg).
Report Question
0%
50
mole percent
0%
52
mole percent
0%
32
mole percent
0%
48
mole percent
Explanation
1 atm = 760 mm Hg =
P
T
P
T
=
P
0
A
X
A
×
P
0
B
X
B
760
=
250
X
A
+
1000
(
1
−
X
A
)
240
=
750
X
A
X
A
=
0.32
Hence mole
%
of A
=
32
%
At 80C, the vapour pressure of pure benzene is 753 mm Hg and of pure toluene 290 mm Hg. Calculate the composition of a liquid in mole percent which at
80
is in equilibrium with vapour containing 30 mole percent of benzene.
Report Question
0%
24, 56
0%
26 ,54
0%
56 24
0%
34 ,76
Explanation
Mole % of benzene=30%of solution
Moles of benzene=30*80/100=24
Moles of Toluene=70*80/100=56
Composition of benzene and Toluene are 24 moles and 56 moles respectively.
Two moles of pure liquid 'A' (
P
o
A
= 80mm of Hg) and 3 moles of pure liquid 'B' (
P
o
B
=
120
mm of
H
g
) are mixed. Assuming ideal behaviour.
Report Question
0%
Vapour pressure of the mixture is
104
mm of
H
g
0%
Mole fraction of liquid '
A
′
in Vapour pressure is
0.3077
0%
Mole fraction of '
B
' in vapour pressure is
0.692
0%
Mole fraction of '
B
' in vapour pressure is
0.785
Explanation
Formula of Ammonia =
N
H
3
P
∘
A
=
80
m
m
H
g
n
A
=
2
mole
P
∘
B
=
120
m
m
H
g
n
B
=
3
mole
x
A
=
n
A
n
A
+
n
B
=
2
2
+
3
=
0.4
x
B
=
n
B
n
A
+
n
B
=
3
2
+
3
=
0.6
Vapour pressure
P
A
=
P
∘
A
x
A
+
P
∘
B
x
B
P
A
=
80
×
0.4
+
120
×
0.6
=
104
m
m
H
g
Option A is correct
Match the column I with column II and mark the appropriate choice
Column I
Column II
(A) Ethyl alcohol + water
(i)
p
=
p
o
x
(B) Benzene + Toulene
(ii) Effect of pressure on gas solutions
(C) Henry's law
(iii) Ideal solution
(D) Raoult's law
(iv) Azeotropic mixture
Report Question
0%
(A)
→
(i); (B)
→
(ii); (C)
→
(iii); (D)
→
(iv)
0%
(A)
→
(i); (B)
→
(iii); (C)
→
(ii); (D)
→
(iv)
0%
(A)
→
(iv); (B)
→
(iii); (C)
→
(ii); (D)
→
(i)
0%
(A)
→
(iii); (B)
→
(i); (C)
→
(i); (D)
→
(iv)
Explanation
Column I
Column II
(A) Ethly alcohol
+
water
−
Azerotropic mixture
(B) Benzene
+
Toluene
−
Ideal solution
(C) Henry's law
−
Effect of pressure on gas solutions
(D) Raolt's law
−
p
=
p
0
x
.
What are the conditions for an ideal solution which obeys Raoult's law over the entire range of concentration?
Report Question
0%
Δ
m
i
x
H
=
0
,
Δ
m
i
x
V
=
0
,
P
T
o
t
a
l
=
p
o
A
x
A
+
p
o
B
x
B
0%
Δ
m
i
x
H
=
+
v
e
,
Δ
m
i
x
V
=
0
,
P
T
o
t
a
l
=
p
o
A
x
A
+
p
o
B
x
B
0%
Δ
m
i
x
H
=
0
,
Δ
m
i
x
V
=
+
v
e
,
P
T
o
t
a
l
=
p
o
A
x
A
+
p
o
B
x
B
0%
Δ
m
i
x
H
=
0
,
Δ
m
i
x
V
=
0
,
P
T
o
t
a
l
=
p
o
B
x
B
Explanation
For an ideal solution that obeys Raoult's Law the following are valid:
Heat is neither released nor absorbed during the reaction,
Δ
m
i
x
H
=
0
The volume of the solution remains the same
Δ
m
i
x
V
=
0
For an ideal solution,
Δ
H
and
Δ
V
for mixing should be zero.
P
T
o
t
a
l
=
p
A
+
p
B
and A-A, B-B and A-B interactions should be nearly the same.
For which of the following solutes the van't Hoff factor is not greater than one?
Report Question
0%
N
a
N
O
3
0%
B
a
C
l
2
0%
K
4
[
F
e
(
C
N
)
6
]
0%
N
H
2
C
O
N
H
2
Explanation
N
H
2
C
O
N
H
2
→
urea
urea is a non-electrolyte and for non-electrolytes
i
=
1
i
=
1
−
α
+
n
α
i
=
1
{
α
=
0
for non-electrolytes}
A solute
X
when dissolved in a solvent associate to form a pentamer. The value of van't Hoff factor (
i
) for the solute will be:
Report Question
0%
0.5
0%
5
0%
0.2
0%
0.1
Explanation
n
A
⟶
(
A
)
n
initially
1
m
o
l
0
after assoociation
1
−
α
α
/
n
i
=
1
−
α
+
α
n
i
=
1
−
1
+
1
n
(
∵
α
=
1
)
i
=
1
5
=
0.2
What will be the correct order of vapour pressure of water, acetone and ether at
30
0
C. Given that among these compounds, water has maximum boiling point and ether has minimum boiling point.
Report Question
0%
Water < Ether < Acetone
0%
Water < Acetone < Ether
0%
Ether < Acetone < Water
0%
Acetone < Ether < Water
Explanation
Higher the boiling point lower will be the vapour pressure. Thus, the correct order of vapour pressure is Water < Acetone < Ether.
If
α
is the degree of dissociation of
N
a
2
S
O
4
, the vant Hoff's factor (
i
) used for calculating the molecular mass is:
Report Question
0%
1
+
α
0%
1
−
α
0%
1
+
2
α
0%
1
−
2
α
Explanation
For
N
a
2
S
O
4
:
−
i
=
n
N
a
2
S
O
4
⟶
2
N
a
+
+
S
O
2
−
4
⏟
n
=
2
i
=
1
−
α
+
n
α
i
=
1
−
α
+
2
α
i
=
1
+
2
α
6
g
. of the area of dissolved in
90
g
. of boiling water. The vapour pressure of the solution is:
Report Question
0%
744.8
mm
0%
758
mm
0%
761
mm
0%
760
mm
Explanation
Vapour pressure
P=P°x,\quad P°=
Boiling point of pure solvent
P=P°\cfrac { { n }_{ 1 } }{ { n }_{ 1 }+{ n }_{ 2 } }
=760\left( \cfrac { 0.1 }{ 0.1+5 } \right) \quad \quad 90g{ H }_{ 2 }O=5mol={ n }_{ 2 };\quad 6g
urea
=0.1mol={ n }_{ 1 }
=760\left( \cfrac { 0.1 }{ 0.1+5 } \right)
=760\times 0.0196=14.89
Torr=lowered pressure
Now, 1 torr = 1 mm of Hg
Vapour pressure of solution
=760-14.89\approx 744.8 mm\ of\ Hg
Option A is correct.
Among the following substances, the lowest vapour pressure is exerted by:
Report Question
0%
water
0%
alcohol
0%
ether
0%
mercury
Explanation
Vapor pressure of mercury is very low (compared to a liquid like methyl alcohol) because the forces of interaction between the individual metal atoms of mercury is quite a bit stronger than the cohesive molecular forces (such as a hydrogen bonding) that holds together several molecules in case of alcohols,ethers and water.
The van't Hoff factor of
0.005M
aqueous solution of
KCl
is
1.95
. The degree of ionisation of
KCl
is:
Report Question
0%
0.95
0%
0.97
0%
0.94
0%
0.96
Explanation
The van't Hoff factor
is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its
mass. For most non-electrolytes dissolved in water, the van't Hoff factor
is essentially 1.
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Practice Class 12 Engineering Chemistry Quiz Questions and Answers
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