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CBSE Questions for Class 11 Medical Chemistry Some Basic Concepts Of Chemistry Quiz 11 - MCQExams.com
CBSE
Class 11 Medical Chemistry
Some Basic Concepts Of Chemistry
Quiz 11
3
g
of a metal
M
combines with
2
g
of oxygen to form its metal oxide.
0.03
equivalent of this metal oxide is dissolved in
500
m
l
of water to form
0.01
molar aqueous solution. Molar mass of metal oxide is (assuming no change in volume of solution)
Report Question
0%
180
g
0%
160
g
0%
140
g
0%
120
g
There are two common oxides of Sulphur, one of which contains
50
%
O
2
by weight, the other almost exactly
60
%. The weights of sulphur which combine with
1
g
of
O
2
(Fixed) are in the ratio of:
Report Question
0%
1
:
1
0%
2
:
1
0%
2
:
3
0%
3
:
2
Explanation
There are two common oxides of sulphur,one of which contains 50%
O
2
by weight the other almost exactly 60%.
60
% of sulphur % of oxygen
First oxide 50% 50%
Second oxide 40% 60%
In first oxide 1 part of oxygen combines with sulphur =
50
50
=
1
Im second oxide 1 part of oxygen will combine with sulphur =
40
60
=
0.67
Therefore weight of sulphur which combines with 1 g of
O
2
are in the ratio :
1
0.67
=
1
:
0.67
=
3
:
2
Hence option D is correct answer.
Iron forms two oxides, in first oxide
56
grams of Iron is found to be combined with
16
gram oxygen and in second oxide
112
gram of Iron is found to be combined with
48
gram oxygen. This data satisfy the law of
Report Question
0%
Conservation of mass
0%
Reciprocal proportion
0%
Multiple proportions
0%
Combining volume
What is the correct order of occurance (% by weight) in air of Ne, Ar and Kr ?
Report Question
0%
Ne > Ar > Kr
0%
Ar > Ne > Kr
0%
Ar > Kr > Ne
0%
Ne > Kr > Ar.
1
mole of oxygen gas weighs ______________
Report Question
0%
3
g
0%
32
g
0%
1
g
0%
6.022
×
10
22
Mole fraction of glycerine
(
C
3
H
5
(
O
H
)
3
)
in a solution of 36 g of water and 46 g of glycerine is :
Report Question
0%
0.46
0%
0.36
0%
0.20
0%
0.40
Explanation
No. of moles of glycerine = 46/92 (where 92 is M.M. of glycerine)
= 0.5moles
No. of moles of water = 36/18 (where 18 is M.M. of water)
= 2moles
So, mole fraction oh glycerine = 0.5/2.5
= 1/5
= 0.20
Hence the correct option is C.
The density of gas is 0.0002136 g/mL. Its vapour density is :-
Report Question
0%
12
0%
24
0%
48
0%
32
At
2
5
o
C
the density of
15
M
H
2
S
O
4
is
1
.8
g
c
m
−
3
.
Thus, mass percentage of
H
2
S
O
4
is aqueous solution is
Report Question
0%
2%
0%
81.6%
0%
18%
0%
1.8%
All homologues of which of the following hydrocarbons have the same percentage composition
Report Question
0%
Alkanes
0%
Alkynes
0%
Alkenes
0%
Alkadienes
The fractional abundance of
C
l
35
in a sample of chlorine containing only
C
l
35
(atomic weight
=
34.9
) and
C
l
37
(atomic weight
=
36.9
) isotopes, is
0.6
. The average mass number of chlorine is _______________.
Report Question
0%
35.7
0%
35.8
0%
18.8
0%
35.77
Explanation
C
1
35
C
1
37
⇒
0.6
(
1
−
0.6
)
=
0.6
0.4
Average atomic mass
=
34.9
×
0.6
+
36.9
×
0.4
1
Average Atomic mass
=
35.7
Option
A
is the answer.
In which of the following sets all the compounds are ionic?
Report Question
0%
B
C
l
3
,
B
e
C
l
2
,
M
g
O
0%
L
i
F
,
M
g
C
l
2
,
N
a
2
O
0%
N
H
4
C
l
,
B
F
3
,
C
O
2
0%
A
g
C
l
,
A
l
C
l
3
,
P
C
l
5
An element
X
have three isotopes
X
20
,
X
21
and
X
22
. The percentage abundance of
X
20
is
90
%
and the average atomic mass of element is
20.18
. The percentage abundance of
X
21
should be ______________.
Report Question
0%
2
%
0%
8
%
0%
10
%
0%
0
%
Explanation
Average atomic mass
=
∑
%
a
b
u
n
d
a
n
c
e
×
A
t
o
m
i
c
m
a
s
s
100
20.18
=
90
×
20
+
x
×
21
+
(
10
−
x
)
22
100
2018
=
1800
+
21
x
+
220
−
22
x
x
=
2020
−
2018
x
=
2
%
Option
(
A
)
.
1.0
g
of
M
g
is burnt with
0.28
g
of
O
2
in a closed vessel. Which reactant is left in excess and how much?
Report Question
0%
M
g
,
5.8
g
0%
M
g
,
0.58
g
0%
O
2
,
0.24
g
0%
O
2
,
2.4
g
Explanation
2
M
g
+
O
2
→
2
M
g
O
∴
1
m
o
l
e
of
O
2
reacts with
2
moles of
M
g
∵
Moles of
O
2
=
0.28
32
=
0.00875
0.00875
m
o
l
e
of
O
2
reacts with
2
1
×
0.00875
m
o
l
e
s
of
M
g
=
0.0715
M
o
l
e
s
of
M
g
Hence, mass of magnesium that reacts
=
m
o
l
e
s
×
m
o
l
a
r
m
a
s
s
=
0.0715
×
24
=
0.42
g
That means, out of the
1
g
of
M
g
, only
0.42
g
is used.
Therefore magnesium is in excess by
(
1
−
0.42
)
=
0.58
g
.
Hence, option
B
is correct.
Percentage of oxygen in urea is about :
Report Question
0%
20.24
%
0%
26.64
%
0%
46.19
%
0%
7.86
%
A sample of clay contains
50
%
silica and
10
%
water. The sample is partially dried by which it loses
8
g
water. If the percentage of silica in the partially dried clay is
52
, what is the percentage of water in the partially dried clay?
Report Question
0%
2.0
%
0%
6.4
%
0%
10.4
%
0%
2.4
%
Explanation
Let weight of clay is
x
g
m
∴
Weight of silica
=
50
100
×
x
=
0.5
x
g
m
∴
Weight of water
=
10
100
×
x
=
0.1
x
g
m
Now,
8
g
m
water is reduced
∴
weight of clay
=
(
x
−
8
)
g
m
Weight of silica
=
52
100
×
(
x
−
8
)
g
m
=
(
0.52
x
−
4.16
)
g
m
Now, Weight of silica in both case are equal.
∴
0.5
x
=
0.52
x
−
4.16
0.02
x
=
4.16
∴
x
=
208
g
m
=
weight of clay initially.
∴
weight of water
=
0.1
x
=
20.8
g
m
Now,
8
g
water is reduced.
∴
Weight of water
=
20.8
−
8
x
=
12.8
g
m
Weight of clay
=
208
−
8
g
m
=
200
g
m
∴
%
of water
=
12.8
200
×
100
=
6.4
%
Compound Q contains
40
%
carbon by mass.
What could Q be?
1
. glucose,
C
6
H
12
O
6
2
. starch,
(
C
6
H
10
O
5
)
n
3
. sucrose,
C
12
H
22
O
11
.
Report Question
0%
1
,
2
and
3
are correct
0%
1
and
2
only are correct
0%
2
and
3
only are correct
0%
1
only is correct
Explanation
percentage mass of C
=
mass of C X 100%
total mass
1. percentage of C in glucose=40%
2. percentage of C in starch=44%
3. percentage of C in sucrose=42%
Solid ammonium dichromate decomposes as under:
(
N
H
4
)
2
C
r
2
O
7
→
N
2
+
C
r
2
O
3
+
4
H
2
O
if 63 g of ammonium dichromate decomposes, calculate
the volume of
N
2
evolved at STP.
Report Question
0%
11.2 L
0%
5.6 L
0%
22.4 L
0%
1 L
0%
44.8 L
A quantity of
13.5
g
of aluminium when change to
A
l
3
+
ion in solution, will lose
(
A
l
=
27
)
.
Report Question
0%
18.0
×
10
23
electrons
0%
6.02
×
10
23
electrons
0%
3.01
×
10
23
electrons
0%
9.0
×
10
23
electrons
Explanation
Mole of
A
l
=
13.5
27
=
0.5
m
o
l
e
A
l
→
A
l
3
+
As there is a loss of
3
e
−
∴
Total no. of
e
−
=
0.5
×
6.02
×
10
23
×
3
=
9.03
×
10
23
e
−
So, the correct option is
D
A certain mixture of
M
n
O
and
M
N
O
2
contains
66.6
m
o
l
per cent of
M
n
O
2
. What is the approximate mass percent of
M
n
in it?
(
M
n
=
55
)
Report Question
0%
66.67
0%
24.02
0%
72.05
0%
69.62
Explanation
Man of
M
n
O
=
66.67
×
70.93
=
4728.9
g
man of
M
n
O
2
=
33.33
×
86.93
=
2897.37
Total man
=
4728.9
+
2897.37
=
7626.2
g
man of
M
n
=
55
×
100
=
5500
%
M
n
=
5500
7626.2
×
100
=
72.11
%
The mass composition of universe may be given as
90
%
H
2
and
10
%
H
e
. The average molecular mass of universe should be ____________.
Report Question
0%
2.20
0%
2.10
0%
3.80
0%
3.64
Explanation
Average M.M of universe
=
0.90
×
2
+
0.10
×
4
0.90
+
0.10
=
1.80
+
0.40
1
=
2.20
Option
B
ic correct.
From
2
m
g
calcium
1.2
×
10
19
atoms are removed. The number of
g
- atoms of calcium left is
(
C
a
=
40
)
:
Report Question
0%
5
×
10
−
5
0%
2
×
10
−
5
0%
3
×
10
−
5
0%
5
×
10
−
6
Explanation
2
m
g
Calcium moles
=
2
×
10
−
3
40
=
5
×
10
−
5
1
mole
=
6.022
×
10
23
atoms
x
=
1.2
×
10
19
6.022
×
10
23
=
1.99
×
10
−
5
moles removed
∴
Remaining moles
=
5
×
10
−
5
−
1.99
×
10
−
5
=
3.00
×
10
−
5
moles
1
g
atom
=
1
mole
∴
Option
C
correct.
In the atomic weight determination. Dalton suggest the formula of water as
H
O
and the composition of water as hydrogen
=
12.5
%
and oxygen
=
87.5
%
by weight. What should be the atomic weight of oxygen on
H
−
scale, on the basis of this information?
Report Question
0%
16
0%
8
0%
14
0%
7
Explanation
Let mass of oxygen be
′
x
′
%
of oxygen in
H
O
87.5
=
m
a
s
s
o
f
o
x
y
g
e
n
m
a
s
s
o
f
c
o
m
p
o
u
n
d
×
100
87.5
=
x
1
+
x
×
100
87.5
(
1
+
x
)
=
100
x
87.5
+
87.5
x
=
100
x
100
x
−
87.5
x
=
87.5
12.5
x
=
87.5
x
x
=
87.5
12.5
x
=
∴
Atomic weight of oxygen
=
7
Option
′
d
′
correct
Dopamine is neurotransmitter, a molecular that serves to transmit message in the brain. The chemical formula of dopamine is
C
8
H
11
O
2
N
. How many moles are there in
1
g
of dopamine?
Report Question
0%
0.00654
0%
153
0%
0.0654
0%
N
o
n
e
o
f
t
h
e
s
e
Explanation
C
8
H
11
O
2
N
Molar mass
=
8
×
12
+
11
×
1
+
2
×
16
+
1
×
14
=
153
g
/
m
o
l
1
m
o
l
=
153
g
x
m
o
l
=
1
g
x
=
1
153
m
o
l
x
=
0.00654
Option
a
A sample of impure cuprous oxide contains
66.67
%
copper, by ,mass. What is the percentage of pure
C
u
2
O
in the sample?
(
C
u
=
63.5
)
Report Question
0%
66.67
0%
75
0%
70
0%
80
Explanation
100
g
m
of sample contains
=
66.6
g
m
mole of
C
u
=
66.6
63.6
=
1.05
m
o
l
e
1
mole of oxygen in
2
mole
C
u
2
mole of
C
u
is with
1
mole oxygen
1.05
mole contains
1
2
×
1.05
weight of oxygen
=
52
×
16
=
84
pure
C
u
2
O
=
66.6
+
8.42
=
75
Molecular mass of dry air is ________________.
Report Question
0%
less than moist air
0%
greater than moist air
0%
equal to moist air
0%
may be greater or less than moist air
Explanation
The molecular mass of dry ice is greater than moist air.
As dry air consists of nitrogen and oxygen while moist air contains water vapour which less mass than that of oxygen and nitrogen making the mass of moist air lower.
M
.
M
of dry air
≃
28.9
g
m
M
.
M
of moist air
<
28.9
g
m
When
2.5
g
of a sample of Mohr's salt reacts completely with
50
m
L
of
N
10
K
M
n
O
4
solution. The % purity of the sample of Mohr's salt is:
Report Question
0%
78.4
0%
70
0%
37
0%
40
Explanation
meq.
F
e
S
O
4
(
N
H
4
)
2
S
O
4
.
6
H
2
O
= meq. of
K
M
n
O
4
(
n
=
1
)
W
392
×
1
×
1000
=
0.1
×
50
;
W
=
1.96
g
Hence,
%
purity of Mohr's salt
=
1.96
2.5
×
100
=
78.4
%
The composition of a sample of wustite is
F
e
0.95
O
1.00
.
What is the percentage of Fe(III) in the sample?
Report Question
0%
10.52%
0%
3.46%
0%
13.84%
0%
93.08%
Explanation
The composition of a sample of wustite is
F
e
0.95
O
. Formula of ferric oxide is
F
e
O
. So,number of
F
e
+
2
ion missing=0.05
Each
F
e
+
2
ion contains +2 charge. So total charge missing =2
×
0.05=0.10.
To maintain electrical neutrality the 0.10 positive charge is compensated by
F
e
+
3
ions.
Replacement of one ferrous ion by one ferric ion increases +1 charge. So no of ferric ions required to compensate the 0.10 charge is 0.10.
So 0.95
F
e
+
2
ions require 0.1
F
e
+
3
ions
So, in 100 Fe atoms, percentage of
F
e
+
3
is =
(
.1
/
.95
)
×
100
=10.52
A vessel contains
1
mole of
O
2
gas at a temperature
T
. The pressure of the gas is
P
. An identical vessel containing
1
mole of
H
e
gas at a temperature
2
T has a pressure of?
Report Question
0%
P
/
8
0%
P
0%
2
P
0%
8
P
Explanation
Moles of
O
2
=
n
1
=
1
T
1
=
T
K
P
1
=
P
atm
Moles of He
=
h
2
=
1
T
2
=
2
T
K
Let pressure be
P
2
V
1
=
V
2
=
V
(given)
P
1
V
1
=
n
2
T
1
∴
P
V
=
(
1
)
(
R
)
(
T
)
…….
(
1
)
∴
P
2
V
=
(
1
)
(
R
)
(
2
T
)
…….
(
2
)
Equation
(
1
)
divided by equation
(
2
)
, we get
∴
P
P
2
=
1
2
∴
P
2
=
2
P
Option C.
A mixture of
H
e
(
4
)
and
N
e
(
20
)
in a
5
−
litre flask at
300
K
and
1
atm weighs
4
g
. The mole
%
of
H
e
is
Report Question
0%
2
0%
0.02
0%
20
0%
4
Explanation
Using
P
V
=
n
R
T
n
=
P
V
R
T
=
1
×
5
0.082
300
=
0.203
m
o
l
e
s
Let the amount of
H
e
in the mixture be
x
, thus amount of
N
e
=
4
−
x
g
Total moles = 0.203
∴
x
4
+
4
−
x
20
=
0.203
x
=
0.012
g
Thus moles of
H
e
in the mixture is
0.012
4
=
0.003
which is 2% moles of the mixture.
What is involved in
C
O
2
and
C
H
4
inclusion?
Report Question
0%
Diamond inclusion
0%
Fluid inclusion
0%
Natural inclusion
0%
Crystal inclusion
What percentage of hydrogen fluoride molecules is trimerized?
Report Question
0%
40
0%
58.8
0%
76.47
0%
17.65
Explanation
Consider mole of
H
E
=
4
H
3
F
3
=
1
then
H
F
in it
=
1
molecules
H
2
F
2
=
y
then
H
F
in it
=
2
molecules
Total mole
=
4
+
1
+
y
=
5
+
y
X
H
E
=
4
5
+
0
,
X
H
2
F
=
y
5
+
y
,
X
H
3
F
3
=
1
5
+
y
20
(
y
5
+
y
)
+
40
(
4
5
+
y
)
+
60
(
1
5
+
y
)
=
34
80
+
40
y
+
60
=
170
+
94
y
40
y
−
34
y
=
170
−
140
6
y
=
30
y
=
5
Total molecular
m
monomer
=
4
dimmer
=
5
×
2
=
10
trimer
=
1
×
3
=
3
Total molecules
=
4
+
10
+
3
=
17
(
5
)
%
trimer
=
3
17
×
10
=
17.64
%
Option
(
d
)
10 grams of a solute is dissolved in 90 grams of a solvent. Its mass percent in solution is
Report Question
0%
0.01
0%
11.1
0%
10
0%
9
Explanation
%
by wt.
=
w
t
.
o
f
t
h
e
s
o
l
u
t
e
(
g
)
w
t
.
o
f
t
h
e
s
o
l
u
t
i
o
n
(
g
)
×
100
%
by wt
=
10
90
+
10
×
100
=
10
%
One of the statements of Dalton's atomic theory is given below:
Compounds are formed when atoms of different elements combine in a fixed ratio.
Which of the following laws is/are not related to this statement?
Report Question
0%
Law of conservation of mass
0%
Law of definite proportions
0%
Law of multiple proportions
0%
Avogadro law
Explanation
Law of conservation of mass states that matter can neither be created nor be destroyed.
Avogadro proposed that equal volumes of gases at the same temperature and pressure should contain equal number of molecules.
Both of the above laws are not related to the above statement.
Which of the following reactions is not according to the law of conservation of mass?
Report Question
0%
2
M
g
(
s
)
+
O
2
(
g
)
→
2
M
g
O
(
s
)
0%
C
3
H
3
(
g
)
+
O
2
(
g
)
→
C
O
2
(
g
)
+
H
2
O
(
g
)
0%
P
4
(
s
)
+
5
O
2
(
g
)
→
P
4
O
10
(
s
)
0%
C
H
4
(
g
)
+
2
O
2
(
g
)
→
C
O
2
(
g
)
+
2
H
2
O
(
g
)
Explanation
Number of atoms in the reactant side is not equal to the number of atoms in the product side for the reaction given in option B.
Mass is measure of the amount of matter.
Report Question
0%
True
0%
False
Explanation
Statement given is true, Mass is defined as amount of matter.
The accepted unit of atomic and molecular mass is:
Report Question
0%
kilogram
0%
gram
0%
pound
0%
atomic mass unit
Explanation
Accepted as unit of atomic and molecular mass is atomic mass unit (u).
Which of the given laws of chemical combination is satisfied by the figure?
Report Question
0%
Law of multiple proportion
0%
Gay Lussac's law
0%
Avogadro law
0%
Law of definite proportion
"Equal volumes of all gases at the same temperature and pressure contain equal number of particles." This statement is a direct consequence of:
Report Question
0%
Avogadro's law
0%
Charles law
0%
Ideal gas equation
0%
Law of partial pressure
Explanation
Avogadro's Law states that
"Equal volumes of all gases at the same temperature and pressure contain equal number of particles."
Hence, Option "A" is the correct answer.
40
m
L
gaseous mixture of
C
O
,
C
H
4
and Ne was exploded with
10
m
L
of oxygen. On cooling the gases occupied
36.5
m
L
. After treatment with KOH the volume reduced by
9
m
L
again on treatment with
alkaline pyrogallol, the volume further reduced.
Percentage of
C
H
4
in the original mixture is:
Report Question
0%
22.5
0%
77.5
0%
7.5
0%
15
Explanation
Let the volume of
C
O
,
C
H
4
and
N
e
be
x
,
y
and
z
respectively
C
O
+
1
2
O
2
→
C
O
2
;
x
x
/
2
x
C
H
4
+
2
O
2
→
C
O
2
+
2
H
2
O
(
l
)
y
2
y
y
remaining volume of
O
2
=
10
−
x
2
−
2
y
Volume after reaction :
x
+
y
+
10
−
x
2
−
2
y
+
z
=
36.5
............(i)
x
+
y
=
9
..........(ii)
x
+
y
+
z
=
40
...............(iii)
by Eq. (i) , (ii) & (iii)
Volume of
C
H
4
=
6
m
L
%
of
C
H
4
=
6
40
×
100
⇒
15
How many molecules are there in one mole of a compound?
Report Question
0%
60.22
×
10
−
23
0%
6.022
×
10
−
23
0%
60.22
×
10
23
0%
6.022
×
10
23
4
mole of a mixture of Mohr's salt and
F
e
2
(
S
O
4
)
3
requires
500
m
L
of
1
M
K
2
C
r
2
O
7
for complete oxidation in acidic medium. The mole % of the Mohr's salt in the mixture is:
Report Question
0%
25
0%
50
0%
60
0%
75
Explanation
C
r
2
O
2
−
7
+
6
F
e
2
+
+
14
H
+
→
2
C
r
3
+
+
6
F
e
3
+
+
7
H
2
O
(
n
=
1
)
(Mohr's salt)
Equivalent of
F
e
2
+
= moles of Mohr's salt
= equivalent of
K
2
C
r
2
O
7
=
500
×
10
−
3
×
6
×
1
=
3.0
Hence. mole percent of Mohr's salt
=
3
4
×
100
=
75
The weight percentage of NaCl solution isIf the weight of the solution is 150 gms, then calculate the weight of Na and water in gms.
Report Question
0%
15,135
0%
25,125
0%
50,100
0%
75,75
Explanation
Weight percent
=
w
e
i
g
h
t
o
f
s
o
l
u
t
e
w
e
i
g
h
t
o
f
s
o
l
u
t
i
o
n
×
100
%
Given, Weight percent
=
10
Weight of solution
=
150
g
m
s
⇒
10
=
w
e
i
g
h
t
o
f
s
o
l
u
t
e
150
g
m
s
×
100
%
∴
Weight of solute
=
15
g
m
s
∴
Weight of water
=
(
150
−
15
)
=
135
g
m
s
.
5
g of crystalline salt, when rendered anhydrous, lost
1.89
g of water. The formula weight of anhydrous salts is
160
. The number of molecules of water of crystallisation in the salt is:
Report Question
0%
1
0%
2
0%
3
0%
5
Explanation
Given,
1.89
g of water is present in
5
g of hydrated salt.
Therefore, weight of anhydrous salt is
(
5
−
1.89
)
g
=
3.11
g.
No of moles of anhydrous salt
=
3.11
160
=
0.02
m
o
l
e
s
No of moles of water
=
1.89
18
=
0.1
m
o
l
e
s
Ratio of anhydrous salt to water,
Salt: Water
=
0.02
:
0.1
1
:
5
Therefore, no of molecules of water
=
5
.
N
a
2
S
O
x
H
2
O
has 50%
H
2
O
by mass. Hence,
x
is:
Report Question
0%
4
0%
45
0%
6
0%
7
Explanation
N
a
2
S
O
3
.
x
H
2
O
has 50%
H
2
O
.
∴
18
x
126
+
18
x
=
50
100
∴
x
=
7
Given that the abundances of isotopes
54
F
e
,
56
F
e
and
57
F
e
are
5
%
,
90
%
and
5
%
respectively, the atomic mass of
F
e
is:
Report Question
0%
55.85
0%
55.95
0%
55.75
0%
56.05
Explanation
ˉ
A
=
∑
A
i
x
i
∑
x
i
ˉ
A
=
54
×
0.05
+
56
×
0.90
+
57
×
0.05
ˉ
A
=
55.95
In a gaseous mixture, if an alkane
(
C
x
H
2
x
+
2
)
and an alkene
(
C
y
H
2
y
)
are taken in
2
:
1
mole ratio, the average molecular weight of the mixture is observed to be
20
. If the same alkane and alkene are taken in
1
:
2
mole ratio, the average molecular weight of the mixture is observed to be
24
. Then, the value of
x
and
y
are respectively :
Report Question
0%
2
,
1
0%
1
,
2
0%
2
,
3
0%
3
,
2
Explanation
If the alkane,
C
x
H
2
x
+
2
and alkene,
C
y
H
2
y
is in the ratio of
2
:
1
, then
M
m
i
x
=
2.
(
14
x
+
2
)
+
1.
(
14
y
)
3
=
20
28
x
+
14
y
=
56
...( 1 )
If the alkane,
C
x
H
2
x
+
2
and alkene,
C
y
H
2
y
is in the ratio of
1
:
2
, then
M
m
i
x
=
1.
(
14
x
+
2
)
+
2.
(
14
y
)
3
=
24
14
x
+
28
y
=
70
...( 2 )
Solving equation (1) and (2), we get
x
=
1
,
y
=
2
In what ratio should a 15% solution of acetic acid be mixed with a 3% solution of the acid to prepare a 10% solution?
[All percentages are mass/mass percentages.]
Report Question
0%
7:3
0%
5:7
0%
7:5
0%
7:10
C
u
2
S
and
M
2
S
are isomorphous in which percentage of sulphur is
20.14
% and
12.94
% respectively. The atomic weight of M is:
[Atomic wt. of Cu = 63.5]
Report Question
0%
208
0%
108
0%
112
0%
106
Explanation
Assume atomic weight of
S
and
M
to be
x
and
y
respectively.
Percentage of sulfur is
20.14
% and
12.94
% in
C
u
2
S
and
M
2
S
respectively.
For
C
u
2
S
, 2 moles Cu = 1 mole S.
1g of
C
u
2
S
contains 0.7986g Cu and 0.2014g S respectively.
Hence,
2
×
0.7986
63.5
=
1
×
0.2014
x
1g of
M
2
S
contains 0.8706g M and 0.1294g S respectively.
Hence,
2
×
0.8706
y
=
1
×
0.1294
x
Solving tow equation we get,
y
=
107.70
≈
108.
A 1.24 M aqueous solution of KI has density of
1.15
g
/
c
m
3
. What is the percentage composition of solute in the solution?
Report Question
0%
17.89
0%
27.89
0%
37.89
0%
47.89
Explanation
1.24 M aqueous solution of KI means 1.24 moles of KI in 1 litre of solution.
Given,
Density of solution
=
1.15
g
c
m
3
Volume of solution
=
1
L
o
r
1000
c
m
3
∴
Mass of solution
=
D
e
n
s
i
t
y
×
v
o
l
u
m
e
=
1.15
×
1000
=
1150
g
Mass of solute
=
1.24
×
166
=
205.8
g
Percentage composition of solute
=
Mass of solute
Mass of solution
×
100
=
205.8
1150
×
100
=
17.89
%
Hence, option A is correct.
4.9
g sample of
K
C
l
O
3
was heated under such conditions that a part of it decomposed according to the equation:
2
K
C
l
O
3
→
2
K
C
l
+
3
O
2
and remaining underwent change according to the equation:
4
K
C
l
O
3
→
3
K
C
l
O
4
+
K
C
l
. If the amount of
O
2
evolved was
672
mL at
1
atm and
273
K, the percentage by weight of
K
C
l
O
4
in the residue is :
Report Question
0%
52.72
%
0%
46.64
%
0%
54.86
%
0%
42.35
%
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Practice Class 11 Medical Chemistry Quiz Questions and Answers
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