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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
∴
of
O_{2}
reacts with
2
moles of
Mg
\because
Moles of
O_{2} = \dfrac {0.28}{32} = 0.00875
0.00875\ mole
of
O_{2}
reacts with
\dfrac {2}{1} \times 0.00875\ moles
of
Mg
= 0.0715\ Moles
of
Mg
Hence, mass of magnesium that reacts
= moles\times molar\ mass
= 0.0715\times 24 = 0.42\ g
That means, out of the
1\ g
of
Mg
, 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\ gm
\therefore \
Weight of silica
=\dfrac {50}{100}\times x
=0.5\ x\ gm
\therefore \
Weight of water
=\dfrac {10}{100}\times x=0.1\ x\ gm
Now,
8\ gm
water is reduced
\therefore \
weight of clay
=(x-8)\ gm
Weight of silica
=\dfrac {52}{100}\times (x-8)\ gm
=(0.52 \ x-4.16)\ gm
Now, Weight of silica in both case are equal.
\therefore \ 0.5\ x=0.52\ x-4.16
0.02\ x=4.16
\therefore \ x=208\ gm\ =
weight of clay initially.
\therefore \
weight of water
=0.1\ x=20.8\ gm
Now,
8\ g
water is reduced.
\therefore \
Weight of water
=20.8-8\ x =12.8\ gm
Weight of clay
=208-8\ gm=200\ gm
\therefore \ \%
of water
=\dfrac {12.8}{200}\times 100=6.4\%
Compound Q contains
40\%
carbon by mass.
What could Q be?
1
. glucose,
C_6H_{12}O_6
2
. starch,
(C_6H_{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
\text{percentage mass of C}=\dfrac{\text{mass of C X 100%}}{\text{total mass}}
\text{1. percentage of C in glucose=40%}
\text{2. percentage of C in starch=44%}
\text{3. percentage of C in sucrose=42%}
Solid ammonium dichromate decomposes as under:
( NH_4)_2Cr_2O_7 \rightarrow N_2 + Cr_2O_3 +4H_2O
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
Al^{3+}
ion in solution, will lose
(Al=27)
.
Report Question
0%
18.0\times 10^{23}
electrons
0%
6.02\times 10^{23}
electrons
0%
3.01\times 10^{23}
electrons
0%
9.0\times 10^{23}
electrons
Explanation
Mole of
Al=\dfrac {13.5}{27}=0.5\ mole
Al\to Al^{3+}
As there is a loss of
3e^-
\therefore \
Total no. of
e^-=0.5\times 6.02\times 10^{23}\times 3
=9.03\times 10^{23}\ e^-
So, the correct option is
D
A certain mixture of
MnO
and
MNO_2
contains
66.6\ mol
per cent of
MnO_2
. What is the approximate mass percent of
Mn
in it?
(Mn=55)
Report Question
0%
66.67
0%
24.02
0%
72.05
0%
69.62
Explanation
Man of
MnO=66.67\times 70.93
=4728.9\ g
man of
MnO_2=33.33\times 86.93
=2897.37
Total man
=4728.9+2897.37
=7626.2\ g
man of
Mn=55\times 100=5500
\%Mn=\dfrac {5500}{7626.2}\times 100
=72.11\%
The mass composition of universe may be given as
90\%\ H_2
and
10\%\ He
. 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
=\dfrac{0.90 \times 2 + 0.10 \times4}{0.90+0.10}
=\dfrac{1.80+0.40}{1}=2.20
Option
B
ic correct.
From
\ 2 \ mg
calcium
1.2\times 10^{19}
atoms are removed. The number of
g
- atoms of calcium left is
(Ca=40)
:
Report Question
0%
5\times 10^{-5}
0%
2\times 10^{-5}
0%
3\times 10^{-5}
0%
5\times 10^{-6}
Explanation
2\ mg
Calcium moles
=\dfrac{2\times 10^{-3}}{40}=5\times 10^{-5}
1
mole
=6.022\times 10^{23}
atoms
x=\dfrac{1.2\times 10^{19}}{6.022\times 10^{23}}
=1.99\times 10^{-5}
moles removed
\therefore
Remaining moles
=5\times 10^{-5}-1.99\times 10^{-5}
=3.00\times 10^{-5}
moles
1\ g
atom
=1
mole
\therefore
Option
C
correct.
In the atomic weight determination. Dalton suggest the formula of water as
HO
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
HO
87.5=\dfrac {mass\ of\ oxygen}{mass\ of\ compound}\times 100
87.5=\dfrac {x}{1+x}\times 100
87.5(1+x)=100x
87.5+87.5x=100x
100x-87.5x=87.5
12.5x=87.5x
x=\dfrac {87.5}{12.5}
\boxed {x=}
\therefore \
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%
None\ of\ these
Explanation
C_{8}H_{11}O_{2}N
Molar mass
=8\times 12+11\times 1+2\times 16+1\times 14
=153\ g/mol
1\ mol=153\ g
x\ mol=1\ g
x=\dfrac{1}{153} mol
\boxed{x=0.00654}
Option
a
A sample of impure cuprous oxide contains
66.67\%
copper, by ,mass. What is the percentage of pure
Cu_2O
in the sample?
(Cu=63.5)
Report Question
0%
66.67
0%
75
0%
70
0%
80
Explanation
100\ gm
of sample contains
=66.6\ gm
mole of
Cu=\dfrac {66.6}{63.6}=1.05\ mole
1
mole of oxygen in
2
mole
Cu
2
mole of
Cu
is with
1
mole oxygen
1.05
mole contains
\dfrac {1}{2}\times 1.05
weight of oxygen
=52\times 16=84
pure
Cu_2O=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
\simeq 28.9\ gm
M.M
of moist air
< 28.9\ gm
When
2.5\,g
of a sample of Mohr's salt reacts completely with
50\,mL
of
\dfrac{N}{10} KMnO_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.
FeSO_4 (NH_4)_2SO_4. 6H_2O
= meq. of
KMnO_4
(n = 1)
\dfrac{W}{392} \times 1 \times 1000 = 0.1 \times 50
;
W = 1.96\,g
Hence,
\%
purity of Mohr's salt
= \dfrac{1.96}{2.5} \times 100 = 78.4 \%
The composition of a sample of wustite is
Fe_{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
Fe_{0.95}O
. Formula of ferric oxide is
FeO
. So,number of
Fe^{+2}
ion missing=0.05
Each
Fe^{+2}
ion contains +2 charge. So total charge missing =2
\times
0.05=0.10.
To maintain electrical neutrality the 0.10 positive charge is compensated by
Fe^{+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
Fe^{+2}
ions require 0.1
Fe^{+3}
ions
So, in 100 Fe atoms, percentage of
Fe^{+3}
is =
(.1/.95)\times 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
He
gas at a temperature
2
T has a pressure of?
Report Question
0%
P/8
0%
P
0%
2P
0%
8P
Explanation
Moles of
O_2=n_1=1
T_1=T
K
P_1=P
atm
Moles of He
=h_2=1
T_2=2T
K
Let pressure be
P_2
V_1=V_2=V
(given)
P_1V_1=n^2T_1
\therefore PV=(1)(R)(T)
…….
(1)
\therefore P_2V=(1)(R)(2T)
…….
(2)
Equation
(1)
divided by equation
(2)
, we get
\therefore \dfrac{P}{P_2}=\dfrac{1}{2}
\therefore P_2=2P
Option C.
A mixture of
He(4)
and
Ne(20)
in a
5-
litre flask at
300\ K
and
1
atm weighs
4\ g
. The mole
\%
of
He
is
Report Question
0%
2
0%
0.02
0%
20
0%
4
Explanation
Using
PV=nRT
n= \dfrac{PV}{RT} = \dfrac{1 \times 5}{0.082}{300} =0.203 \ moles
Let the amount of
He
in the mixture be
x
, thus amount of
Ne
=
4-x
g
Total moles = 0.203
\therefore \dfrac{x}{4} + \dfrac{4-x}{20} = 0.203
x= 0.012 \ g
Thus moles of
He
in the mixture is
\dfrac{0.012}{4} = 0.003
which is 2% moles of the mixture.
What is involved in
CO_2
and
CH_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
HE=4
H_{3}F_{3}=1
then
HF
in it
=1
molecules
H_{2}F_{2}=y
then
HF
in it
=2
molecules
Total mole
=4+1+y=5+y
X_{HE}=\dfrac{4}{5+0}, X_{H_{2}F}=\dfrac{y}{5+y}, X_{H_{3}F_{3}}=\dfrac{1}{5+y}
20\left(\dfrac{y}{5+y}\right)+40\left(\dfrac{4}{5+y}\right)+60\left(\dfrac{1}{5+y}\right)=34
80+40y+60=170+94 y
40y-34y=170-140
6y=30
y=5
Total molecular
m
monomer
=4
dimmer
=5\times 2=10
trimer
=1\times 3=3
Total molecules
=4+10+3=17
(5)\ \%
trimer
=\dfrac{3}{17}\times 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.
= \dfrac {wt. \,of \,the \,solute (g)}{wt. \,of \,the \,solution(g)} \times 100
\%
by wt
= \dfrac {10}{90 + 10} \times 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%
2Mg(s)+O_2(g)\to 2MgO(s)
0%
C_3H_3(g)+O_2(g)\to CO_2(g)+H_2O(g)
0%
P_4(s)+5O_2(g)\to P_4O_{10}(s)
0%
CH_4(g)+2O_2(g)\to CO_2(g)+2H_2O(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\,mL
gaseous mixture of
CO , CH_4
and Ne was exploded with
10\,mL
of oxygen. On cooling the gases occupied
36.5\,mL
. After treatment with KOH the volume reduced by
9\,mL
again on treatment with
alkaline pyrogallol, the volume further reduced.
Percentage of
CH_{4}
in the original mixture is:
Report Question
0%
22.5
0%
77.5
0%
7.5
0%
15
Explanation
Let the volume of
CO , CH_4
and
Ne
be
x , y
and
z
respectively
CO + \dfrac{1}{2} O_2 \rightarrow CO_2
;
x
x/2
x
CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O (l)
y
2y
y
remaining volume of
O_2 = 10 - \dfrac{x}{2}- 2y
Volume after reaction :
x + y + 10 - \dfrac{x}{2} - 2y + z = 36.5
............(i)
x + y = 9
..........(ii)
x + y + z = 40
...............(iii)
by Eq. (i) , (ii) & (iii)
Volume of
CH_4 = 6\,mL
\%
of
CH_4 = \dfrac{6}{40} \times 100 \Rightarrow 15
How many molecules are there in one mole of a compound?
Report Question
0%
60.22\times 10^{-23}
0%
6.022\times 10^{-23}
0%
60.22\times 10^{23}
0%
6.022\times 10^{23}
4
mole of a mixture of Mohr's salt and
Fe_2(SO_4)_3
requires
500\,mL
of
1\,M \,K_2Cr_2O_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
Cr_2O_{7}^{2-} + 6Fe^{2+} + 14 H^+ \rightarrow 2Cr^{3+} + 6Fe{3+} + 7H_2O
( n = 1 )
(Mohr's salt)
Equivalent of
Fe^{2+}
= moles of Mohr's salt
= equivalent of
K_2Cr_2O_7
=
500 \times 10^{-3} \times 6 \times 1 = 3.0
Hence. mole percent of Mohr's salt
= \dfrac{3}{4} \times 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
=\cfrac{weight \,of\, solute}{weight\, of\,solution}\times 100\%
Given, Weight percent
=10
Weight of solution
=150\,gms
\Rightarrow\, 10=\cfrac{weight\,of\,solute}{150\,gms}\times100\%
\therefore
Weight of solute
=15\,gms
\therefore
Weight of water
=(150-15)=135\,gms
.
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
= \dfrac{3.11}{160} = 0.02 moles
No of moles of water
= \dfrac{1.89}{18} = 0.1 moles
Ratio of anhydrous salt to water,
Salt: Water
= 0.02 : 0.1
1:5
Therefore, no of molecules of water
=5
.
Na_2SO_xH_2O
has 50%
H_2O
by mass. Hence,
x
is:
Report Question
0%
4
0%
45
0%
6
0%
7
Explanation
Na_{2}SO_{3}.x H_{2}O
has 50%
H_{2}O
.
\therefore \dfrac{18x}{126+18x}=\dfrac{50}{100}
\therefore x=7
Given that the abundances of isotopes
^{54}Fe, ^{56}Fe
and
^{57}Fe
are
5\%, 90\%
and
5\%
respectively, the atomic mass of
Fe
is:
Report Question
0%
55.85
0%
55.95
0%
55.75
0%
56.05
Explanation
\bar{A}=\dfrac{\sum A_{i}x_{i}}{\sum x_{i}}
\bar{A}=54\times 0.05+56\times 0.90+57\times 0.05
\bar{A}=55.95
In a gaseous mixture, if an alkane
(C_xH_{2x+ 2})
and an alkene
(C_yH_{2y})
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_xH_{2x+2}
and alkene,
C_yH_{2y}
is in the ratio of
2:1
, then
M_{mix}=\dfrac{2.(14x + 2) + 1.(14y)}{3}=20
28x + 14y = 56
...( 1 )
If the alkane,
C_xH_{2x+2}
and alkene,
C_yH_{2y}
is in the ratio of
1:2
, then
M_{mix}=\dfrac{1.(14x + 2) + 2.(14y)}{3}=24
14x + 28y = 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
Cu_{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
Cu_2S
and
M_2S
respectively.
For
Cu_2S
, 2 moles Cu = 1 mole S.
1g of
Cu_2S
contains 0.7986g Cu and 0.2014g S respectively.
Hence,
2\times \dfrac {0.7986}{63.5} =1 \times \dfrac {0.2014 }{x}
1g of
M_2S
contains 0.8706g M and 0.1294g S respectively.
Hence,
2 \times \dfrac {0.8706}{y} =1 \times \dfrac {0.1294}{x}
Solving tow equation we get,
y=107.70 \approx 108.
A 1.24 M aqueous solution of KI has density of
1.15 g / cm^{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\ \dfrac{g}{cm^3}
Volume of solution
= 1\ L\ or\ 1000\ cm^3
\therefore
Mass of solution
= Density\times volume
= 1.15\times 1000
= 1150\ g
Mass of solute
= 1.24\times 166 = 205.8\ g
Percentage composition of solute
= \dfrac{\text{Mass of solute}}{\text{ Mass of solution}}\times 100
= \dfrac{205.8}{1150}\times 100
= 17.89\%
Hence, option A is correct.
4.9
g sample of
KClO_{3}
was heated under such conditions that a part of it decomposed according to the equation:
2KClO_{3}\rightarrow 2KCl+3O_{2}
and remaining underwent change according to the equation:
4KClO_{3}\rightarrow 3KClO_{4}+KCl
. If the amount of
O_{2}
evolved was
672
mL at
1
atm and
273
K, the percentage by weight of
KClO_{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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