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CBSE Questions for Class 11 Engineering Chemistry States Of Matter Quiz 13 - MCQExams.com
CBSE
Class 11 Engineering Chemistry
States Of Matter
Quiz 13
The solubility of a specific non-volatile salt is 4g in 100 g of water at
25
0
C. If 2.0g, 4.0g and 6.0g of the salt added of 100g of water at
25
0
C, in system X, Y and Z. The vapour pressure would be in the order:
Report Question
0%
Z
>
Y
>
X
0%
X
>
Y
>
Z
0%
Z
>
X
=
Y
0%
X
>
Y
=
Z
The pressure exerted by a mass of x mg resting on the area of 1.00
c
m
2
is 1.00 Pa, then x is
Report Question
0%
10.6
0%
10.3
0%
103
0%
10.2
According to kinetic theory of gases :
Report Question
0%
collisions are always elastic
0%
between collisions, the molecules move in straight lines with constant velocities
0%
only a small number of molecules have a very high velocity
0%
at of the above
200
m
l
of
H
e
at
0.66
a
t
m
pressure and
400
m
l
of
O
2
at
0.52
a
t
m
pressure are mixed in a
400
m
l
vessel at
25
∘
C
. The partial pressure of
H
e
and
O
2
will be:
Report Question
0%
0.33
a
n
d
0.56
0%
0.33
a
n
d
0.52
0%
0.38
a
n
d
0.56
0%
0.25
a
n
d
0.45
Explanation
Partial pressure is the pressure of individual gas when it is placed in a required vessel or in final vessel.
Given :- Initially volume of
H
e
is
200
m
l
at
0.66
a
t
m
after placing it in a vessel of volume
400
m
l
P
1
V
1
=
P
2
V
2
.66
×
200
=
P
2
×
400
P
2
=
0.33
a
t
m
----------- (Partial pressure of
H
e
)
Initially
O
2
is present at
0.25
a
t
m
in
400
m
l
volume and then it is placed in
400
m
l
vessel hence the partial pressure of
O
2
is
0.52
a
t
m
Partial pressure of
H
e
=
0.33
a
t
m
Partial pressure of
O
2
=
0.52
a
t
m
The reduction potential of hydrogen electrode will be positive if
Report Question
0%
P
H
2
=
2
a
t
m
;
[
H
+
]
=
1
M
0%
P
H
2
=
2.5
a
t
m
;
[
H
+
]
=
1.5
M
0%
P
H
2
=
2.5
a
t
m
;
[
H
+
]
=
1
M
0%
P
H
2
=
1
a
t
m
;
[
H
+
]
=
2
M
The vapour pressure of pure water at
25
∘
C
is
30
m
m
. The vapour pressure of
10
%
(
W
/
W
)
glucose solution at
25
∘
C
is :
Report Question
0%
31.5
m
m
0%
30.6
m
m
0%
29.67
m
m
0%
26.56
m
m
Explanation
10
% solution
⇒
10
g
of glucose ,
90
g
of water in
100
g
m
of solution
So, number of moles of water in
90
g
m
=
90
18
=
5
moles
Number of moles of glucose=
10
180
=
0.056
moles
So, total moles=
5
+
0.056
=
5.056
moles
Mole fraction of water=
5
/
5.056
=
0.988
So, final vapour pressure=
P
×
X
w
a
t
e
r
=
30
×
0.980
=
29.6
m
m
Hence, the correct option is
C
A reaction follows the given concentration (M)- time graph. The rate for the reaction at 20 seconds will be
Report Question
0%
410 MS
0%
810 MS
0%
210 MS
0%
710 MS
5.40 gm of an unknown gas at
2
7
o
C
occupies the same volume as 0.14 gm of hydrogen at
1
7
o
C
and same pressure. The molecular weight of unknown gas is
Report Question
0%
79.8
0%
81
0%
79.2
0%
83
At point P and Q,the real gas deviation with respect to ideal gas is respectively
Report Question
0%
Positive,negative
0%
Positive, positive
0%
Negative,Positive
0%
Negative,Negative
Explanation
Z
=
P
V
R
T
at point
Z
=
V
s
e
a
l
V
s
e
a
l
=
V
2
V
1
as from graph
v
2
>
V
1
Z
>
1
=
+
v
E
at point
Q
Z
=
P
r
e
a
l
P
i
d
e
a
=
P
1
P
2
But
P
1
<
P
2
The vapour pressure of pure benzene at a certain temperature is 200 mm Hg. At the same temperature the vapour pressure of a solution containing 2 g of non volatile non electrolyte solid in 78 g of benzene is 195 mm Hg. What is the molecular mass of solid?
Report Question
0%
50
0%
70
0%
85
0%
80
n
moles of Helium gas are placed in a vessel of volume
V
Litre. At T K, if
V
1
is free volume of Helium then diameter of He atom is
Report Question
0%
[
3
2
V
1
π
N
A
n
]
1
/
3
0%
[
3
2
(
V
−
V
1
)
π
N
A
n
]
1
/
3
0%
[
6
(
V
−
V
1
)
π
N
A
n
]
1
/
3
0%
[
6
V
1
π
N
A
n
]
1
/
3
Consider a binary mixture of volatile liquids. If at
X
A
=
0.4
the vapour pressure of solution is
580
t
o
r
r
then the mixture could be
(
p
0
A
=
300
t
o
r
r
,
p
0
B
=
800
t
o
r
r
)
Report Question
0%
C
H
C
l
3
−
C
H
3
C
O
C
H
3
0%
C
6
H
5
C
l
−
C
6
H
5
B
r
0%
C
6
H
6
−
C
6
H
5
C
H
3
0%
n
C
6
H
14
−
n
C
7
H
16
The ratio of densities of a gas at two conditions of temperature and pressure is given by:
Report Question
0%
ρ
2
ρ
1
=
(
p
2
p
1
)
(
T
2
T
1
)
0%
ρ
2
ρ
1
=
(
p
2
p
1
)
(
T
1
T
2
)
0%
ρ
2
ρ
1
=
(
p
1
p
2
)
(
T
1
T
2
)
0%
ρ
2
ρ
1
=
(
p
1
p
2
)
(
T
2
T
1
)
Explanation
The density
(
S
)
of an ideal gas is given as
S
=
P
M
R
T
Where,
P
=
Pressure
T
=
Temperature
R
=
Universal gas content
M
=
Molar Mass
Here,
R
and
M
are constant for a given gas
Therefore;
(refer image 1)
Calculate solubility (in moles / litre) of a saturated aqueous solution of
A
g
3
P
O
4
if the vapour pressure of the solution becomes 750 torr at 373 K
(Assume molality=molarity).
Report Question
0%
2
/
15
0%
1
/
30
0%
10
/
54
0%
20
/
27
18 g of glucose
(
C
6
H
12
O
6
)
is added to 178.2 g of water. The vapour pressure of water for this aqueous solution at
100
∘
C
is ?
Report Question
0%
7.60 Torr
0%
76.00 Torr
0%
752.40 Torr
0%
759.00 Torr
Gay Lussac's law is not valid for which chemical reaction?
Report Question
0%
2
N
O
(
g
)
+
O
2
(
g
)
⟶
2
N
O
2
(
g
)
0%
2
H
I
(
g
)
⟶
H
2
(
g
)
+
I
2
(
g
)
0%
4
B
(
s
)
+
3
O
2
(
g
)
⟶
2
B
2
O
3
(
s
)
0%
2
S
O
2
(
g
)
+
O
2
(
g
)
⟶
2
S
O
3
(
g
)
Which of the following attraction is strongest?
Report Question
0%
0%
0%
0%
Liquids
A
and
B
form an ideal solution. At
30
o
C, the total vapour pressure of a solution containing
1
mol of
A
and
2
mols of
B
is
250
mm
H
g
. The total vapour pressure becomes
300
mm
H
g
when
1
more mol of A is added to the first solution. The vapour pressures of pure
A
and
B
at the same temperature are
Report Question
0%
450
,
150
mm
H
g
0%
150
,
450
mm
H
g
0%
250
,
300
mm
H
g
0%
125
,
150
mm
H
g
The equilibrium constant for the following equilibrium is given at
0
0
C
:
N
a
2
H
P
O
4
.12
H
2
O
(
s
)
→
N
a
2
H
P
O
4
.7
H
2
O
(
s
)
+
5
H
2
O
(
g
)
;
K
p
=
31.25
×
10
−
13
.
The vapor pressure of water is:
Report Question
0%
2
×
10
−
4
a
t
m
0%
5
×
10
−
4
a
t
m
0%
5
×
10
−
2
a
t
m
0%
5
×
10
−
3
a
t
m
An ideal gas is compressed in a closed container its U?
Report Question
0%
Increases
0%
Decreases
0%
Remains same
0%
Both
(
1
)
&
(
2
)
Explanation
When an ideal gas is compressed in a closed container, its internal energy U increases. Compressing an ideal gas increases its temperature and its internal energy increases since U = f(t) for an ideal gas.
45.4 L of dinitrogen reacted with 22.7 L of dioxygen and 45.4 L of nitrous oxide was formed. The reaction is given below :
2
N
2
(
g
)
+
O
2
(
g
)
.
→
.
2
N
2
O
(
g
)
Which law is being obeyed in this experiment?
Report Question
0%
Gay Lussac's law
0%
Law of definite proportion
0%
Law of multiple proportion
0%
Avogadro's law
Which of the following mixture of gases does obey Dalton's Law of partial pressure ?
Report Question
0%
C
l
2
and
S
O
2
0%
C
O
2
and
H
e
0%
O
2
and
C
O
2
0%
N
2
and
O
2
A mixture of 50.ml of
N
H
3
and 60.0 ml of
O
2
gas react as
4
N
H
3
(
g
)
+
5
O
2
(
g
)
→
4
N
O
+
6
H
2
O
(
g
)
If all the gases are at the same temperature and the reaction continues until one of the gases is completely consumed, what volume of water vapour is produced?
Report Question
0%
48 mL
0%
60.0mL
0%
72 mL
0%
75.0 mL
Two liquids A and B have
P
A
0
:
P
B
0
= 1:3 at a certain temperature. If the mole fraction ratio
x
A
:
x
B
=
1:3, the mole fraction of A in vapour in equilibrium with the solution at the given temperature is -
Report Question
0%
0.1
0%
0.2
0%
0.5
0%
1.0
Calculate the mass of a non - volatile solute (molar mass
40
g
m
o
l
−
1
) which should be dissolved in
114
g
octane to reduce its vapour pressure to
80
%
Report Question
0%
2
g
0%
4
g
0%
8
g
0%
10
g
Initial temperature of an ideal gas is
7
5
o
C
. At what temperature, the sample of neon gas would be heated to double its pressure, if the initial volume of gas is reduced by 15%?
Report Question
0%
31
9
o
C
0%
59
2
o
C
0%
12
8
o
C
0%
6
0
o
C
Explanation
T
1
=
75
o
C
=
75
+
273
=
348
K
Let
V
1
=
V
and
V
2
=
85
V
/
100
V
1
/
T
1
=
V
2
/
T
2
T
2
=
V
2
T
1
V
1
=
85
V
×
348
100
×
V
=
295.8
K
P
1
=
P
P
2
=
2
P
T
1
=
295.8
K
T
2
=
?
P
1
/
T
1
=
P
2
/
T
2
T
2
=
P
2
T
1
P
1
=
2
P
×
295.8
P
=
591.6
K
=
591.6
−
273
.
o
C
=
318.6
o
C
319
o
Which of the following solutions will have the lowest vapour pressure?
Report Question
0%
0.1
M
G
l
u
c
o
s
e
0%
0.1
M
N
a
C
l
0%
0.1
M
B
a
C
l
2
0%
0.1
M
A
l
2
(
S
O
4
)
3
At
40
∘
C
the vapour pressure ( in torr ) of a mixture of methyl alcohol and ethyl alcohol is represented by
P
=
199
x
+
135
(where
x
is the mole fraction of methyl alcohol)
What are the vapour pressures of pure methyl alcohol and pure ethyl alcohol at
40
∘
C
?
Report Question
0%
135
and
254
torr
0%
119
and
135
tor
0%
119
and
254
torr
0%
334
and
135
torr
Explanation
Correct option is D
Vapour Pressure of methyl alcohol and ethyl alcohol solution is represented by
P
=
115
x
A
+
140
. Where
x
A
is the mole fraction of methyl alcohol. The value of
lim
x
A
→
0
P
0
B
x
B
Report Question
0%
255
0%
115
0%
140
0%
135
Calculate the vapour pressure of aqueous 0.1 m glucose solution at 300 K temperature, the vapour pressure of water is 0.03 bar at 300 K temperature.
Report Question
0%
0.5 bar
0%
0.29 bar
0%
0.3 bar
0%
0.03 bar
5.0
L
water placed in a closed room of volume
2.5
×
10
4
L
having temperature
300
K
. If vapour pressure of water is
27.0
m
m
and density is
0.990
g
/
c
m
3
at this temperature, how much water is left in liquid state?
Report Question
0%
3.444
L
0%
4.344
L
0%
4.798
L
0%
1.212
L
At a certain temperature pure liquid A and liquid B have vapour pressures 10 torr and 37 torr respectively. For a certain ideal solution of A and B, the vapour in equilibrium with the liquid has the components A and B in the partial pressure ratio
P
A
:
P
B
=
1
:
7
. What is the mole fraction of A in the solution?
Report Question
0%
0.346
0%
0.654
0%
0.5
0%
None of these
The energy of absorbed by each molecule
(
A
2
)
of a substance is
4.4
×
10
−
19
J
and bond energy per molecule is
4.0
×
10
−
19
J
. The kinetic energy of the molecule per atom will be
Report Question
0%
2.2
×
10
−
19
J
0%
2.0
×
10
−
19
J
0%
4.0
×
10
−
20
J
0%
2.0
×
10
−
20
J
Vapour pressure increase with increase in :
Report Question
0%
Concentration of solution containing non-volatile soulute
0%
Temperature up to boiling point
0%
Temperature upto triple point
0%
Altitude of the concerned place of boiling
The vapour pressure of the solution of
5
g
on non-electrolyte in
100
g
of water at a particular temperature is
2950
N
/
m
2
. The vapour pressure of water is
3000
N
/
m
2
. The mocular mass of the solute is ?
Report Question
0%
54
0%
124
0%
180
0%
340
The kinetic energy of two moles of
N
2
at
27
o
C
is
(
R
=
8.314
J
K
−
1
m
o
l
−
1
)
Report Question
0%
5491.6 J
0%
6491.6 J
0%
7482.6 J
0%
8882.4 J
The mass of a non-volatile non-electrolyte solute (molar mass
=
50
g
m
o
l
−
1
) needed to be dissolved in
114
g
octane to reduce its vapour pressure to
75
%
is:
Report Question
0%
37.5
g
0%
75
g
0%
150
g
0%
50
g
At a certain temperature, the vapour pressure of water is
50
m
m
.
The relative lowering of vapour pressure of a solution containing
36
g
of glucose in
900
g
of water is
Report Question
0%
0.1
0%
0.5
0%
0.004
0%
3.75
The vapour pressure of
1
molal glucose solution at
100
o
C
will be
Report Question
0%
760
m
m
H
g
0%
76.0
m
m
H
g
0%
1
a
t
m
0%
746.32
m
m
H
g
0.5
moles of gas
A
and
x
moles of gas
B
exert a pressure of
200
Pa in a a container of volume
10
m
3
at
1000
K. Given
R
is the gas constant in
J
K
−
1
mol
1
,
x
is:
Report Question
0%
2
R
4
+
12
0%
2
R
4
−
12
0%
4
−
R
2
R
0%
4
+
R
2
R
Explanation
Given that
n
T
=
(
0.5
+
x
)
;
T
=
1000
K
;
V
=
10
m
3
;
P
=
200
P
a
Now,
P
V
=
n
×
R
×
T
200
×
10
=
(
0.5
+
x
)
×
R
×
1000
2
=
(
0.5
+
x
)
R
2
R
=
1
2
+
x
4
R
−
1
=
2
x
4
−
R
2
R
=
x
When a solution containing non-volatile solute is diluted with water,
Report Question
0%
its vapour pressure increases
0%
its osmotic pressure increases
0%
its boiling point increases
0%
its freezing point increases
An ideal solution formed by mixing two liquids 'A' and 'B' find the pressure at which
X
A
=
Y
B
(
P
A
0
=
400
t
o
r
r
,
P
B
0
=
100
t
o
r
r
)
(
X
A
and
Y
B
are mole fraction of A and B in liquid and vapour forms respectively at equilibrium.)
Report Question
0%
250 torr
0%
200 torr
0%
150 torr
0%
125 torr
At
35
o
C
, the vapour pressure of
C
S
2
is
512
m
m
H
g
, and acetone is
344
m
m
H
g
. A solution of
C
S
2
and acetone in, which the mol fraction of
C
S
2
is
0.25
, has a total vapour pressure of
600
m
m
H
g
. Which of the following statements is/are correct:
Report Question
0%
A mixture of
100
m
L
of acetone and
100
m
L
C
S
2
of has a volume is
200
m
L
0%
When acetone and
C
S
2
are mixed at
35
o
C
, heat must be absorbed in order to produce a solution at
35
o
C
0%
When acetone and
C
S
2
are mixed at
35
o
C
, heat is released
0%
There is negative deviation from Raoult's law
Vapour pressure of pure 'A' is 70 mm of Hg at
25
0
C
it forms an ideal solution is with 'B' in which mole fraction of A is 0.If the vapour pressure of the solution is 84 mm is Hg at
25
0
C
, the vapour pressure of pure 'B' at
25
0
C
is :
Report Question
0%
56 mm Hg
0%
70 mm Hg
0%
140 mm Hg
0%
28 mm Hg
A solution is prepared by mixing
8
.5
g
of
C
H
2
C
l
2
and
11.95
g
of
CHC
l
3
. If vapour pressure of
C
H
2
C
l
2
and
CHC
l
3
at
298
K
are
415
and
200
mm
Hg respecetively, then mole fraction of
CHC
l
3
in vapour form is:
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0%
0.162
0%
0.675
0%
0.325
0%
0.486
Two liquids A and B form ideal solution.At 300K,the vapour pressure of solution containing 1 mol of A and 3 mol of B is 550mm Hg.At the same temperature,if one more mole of B is added to this solution,the vapour pressure of the solution increases by 10mm Hg.Determine the vapour pressures of A and B in their pure states.
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0%
563
and
469
0%
500
adn
1250
0%
513
and
494
0%
400
and
600
The volume occupied by 4.4 gram of
C
O
2
at STP is
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0%
2.85 liter
0%
2.24 liter
0%
2.6 liter
0%
2.64 liter
At
80
o
C
, the vapour pressure of pure liquid A is 520 mm Hg and that of pure liquid B is 1000 mm Hg. If a mixture solution of A and B boils at
80
o
C
and 1 atm pressure, the amount of A in the mixture is :
(1 atm=760 mm Hg)
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0%
34 mol%
0%
48 mol%
0%
50 mol%
0%
52mol%
The kinetic energy of
1
m
o
l
e
of gas is equal to-
Report Question
0%
3
2
R
T
0%
3
2
K
T
0%
R
T
2
0%
2
R
3
Explanation
Kinetic enedy of gaseous molecule is calculaed using following formula.
K
E
=
3
2
×
n
R
T
.
Above formula is given for n moles
So for 1 mole
n
=
1
K
E
=
3
2
×
1
×
R
T
.
K
E
=
3
2
×
R
T
.
Hence option A is correct.
Pressure exerted by 2 grams of helium present in a vessel is 1.5 atm. If 4 grams of gas 'X' is introduced into the same vessel keeping the condition constant, the pressure is 2.25 atm. Gas 'x' is
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0%
hydrogen
0%
methane
0%
oxygen
0%
soldier oxide
0:0:1
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