Syllabus – cum – Comp./XI/2013–14
Some bas i c c on c ep t s of c h emi s t ry ( XI )
Review No. 2 ( Solutions)
 Objective Questions
Q1.
What is the empirical formula of vanadium oxide if 2.74 g of metal oxide contains 1.53 g of
metal?
(a) V2O3
Sol.
(c) V2O5
(b) VO
(d) V2O7
(c) Amount of O = 2.74-1.53 = 1.21
Wt
Moles
Simplest
Whole no. ratio
M = 1.53
1.53
 0.03
51
.03
1
.03
12=2
O = 1.21
1.21
 .075
16
.075
 2.5
.03
2.5  2 = 5
= V2O5
Q2.
A compound having the empirical formula (C 3H4O)n has a molar mass of 170  5. The
molecular formula of this compound is
(a) C3H4O
Sol.
(d) n =
(b) C6H8O2
(c) C6H12O3
(d) C9H12O3
Molar mass
Emp .mass
Molar mass
= [(12  3) + (4  1) + (16  1)]n = 170  5
= (36 + 4 + 16) n = 170  5
56 n = 170
n=
170  5
n 3
56
 M.F. = (E.F.)n = (C3H4O)3
= C9H12O3
Q3.
What volume of hydrogen at NTP will be liberated when 3.25g of zinc completely dissolve in
dilute HCl? (At. mass of Zn = 65)
(a) 1.12 litre
Sol.
(b) 11.20 litre
(a) Zn + 2HCl  ZnCl2
+
(c) 2.24 litre
(c) 22.40 litre
H2
1 mole
1 mole
65 g
22.4 L
3.25 g
22.4
 3.25
65
= 1.12 L
Q4.
Sucrose reacts with oxygen to yield CO2 and H2O according to the reaction.
C12H22O11 +12O2  12CO2 + 11H2O.
The number of molecules of CO2 produced per gm of sucrose is
(a) 2.11  1022
(b) 6.02  1022
(c) 1.76  1021
(d) 9.29  1023
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[1]
Syllabus – cum – Comp./XI/2013–14
Sol.
(a) C12H22O11  12CO2
1 mole produces 12 moles of CO2
342 g produces 12  NA of CO2
1 g produces
Q5.
12  6.023  10 23
342
= 2.11  1022 molecules
1.6 g of sulphur was burnt in the air to form SO2. The number of molecules of SO2 introduced
into the air will be
(a) 6.02  1023
Sol.
(b) 3.01  1023
(c) 6.02  1022
(d) 3.01  1022
(d) S + O  SO2
1 mole of S produces 1 mole of SO2
32 g of S produces N A molecules of SO2
1.6 g of S produces
Q6.
6.023  10 23
 1.6 = 3.01  1022 molecules
32
Compute the value of x
x = 9.4 g of phenol (C6H5OH) + 6.02 × 1023 (wrong in review) molecules of phenol -0.2 mole of
phenol
(a) 0.9 mol
Sol.
(a)
(b) 9.2 g
(c) 0.1 mol
(d) 6.02 × 1023 molecules
x = 9.4 g of phenol + 6.02  1023 molecules - .2 mole
x=
9 .4
moles + 1 mole - .2 mole
94
= (.1 + 1 - .2) mole
= .9 mole
Q7.
The number of moles of oxygen in one litre of air containing 21% oxygen by volume, in
standard conditions, is
(a) 0.186 mol
Sol.
(c)
(d) 2.10 mol
100 L of air has 21 L of O2
1 L of air has
Q8.
(c) 0.0093 mol
(b) 0.21 mol
21
= .21 L of O2
100
Volume
moles of O2
(At S.T.P.)
22.4 L
1
.21 L
1
 .21 = .0093 moles
22.4
A boy drinks 500 mL of 9% glucose solution. The number of glucose molecules he has
consumed are [mol. wt. of glucose = 180]
(a) 0.5 × 1023
Sol.
(c)
9%
(c) 1.5 × 1023
(b) 1.0 × 1023

W
V

(d) 2.0 × 1023
w eight of solute
Volume of solution
100 mL glucose solution has 9g glucose
S.C.O. 208 II floor, Sec-36 D Chandigarh, Ph.3012085, Mb.9872079084
[2]
Syllabus – cum – Comp./XI/2013–14
9
 500g = 45 g of glucose.
100
500 mL glucose solution has
Moles of glucose =
45
1
  .25
180 4
= .25  6.023  1023
No. of Molecules
= 1.5  1023
Q9.
List –I
(A)
0.5 mol
(p)
92 g CH3CH2OH
(B)
1.0 mol
(q)
66g CO2
(C)
1.5 mol
(r)
54 g H2O
(D)
2.0 mol
(s)
15 g HCHO
(E)
3.0 mol
(t)
44g CH3CHO
Sol.
Q10.
List – II
Moles
=
Given mass
Molar mass
(p)
92
12  12  16  6
=
2
 (D)
(q)
66
44
=
1.5
 (C)
(r)
54
18
=
3
 (E)
(s)
15
30
=
.5
 (A)
(t)
44
44
=
1
 (B)
In the reaction, Al(s) + 6HCl (aq)  2Al3+(aq) + 6Cl- (aq) + 3H2(g)
(a) 33.6 L H2(g) is produced regardless of temperature and pressure for every mole Al that
reacts
(b) 67.2 L H2 (g) at STP is produced for every mole Al that reacts
(c) 11.2 L H2 (g) at STP is produced for every mole HCl (aq) consumed.
(d) 6 L HCl (aq) is consumed for every 3 L H2 (g) produced.
Sol.
Using mole concept
Q11.
When 1 L of CO2 is heated with graphite, the volume of the gases collected is 1.5 L. Calculate
the number of moles of CO produced at STP.
(a)
Sol.
1
11.2
(c)
(b)
CO2 (g) +
Initial volume
1L
Suppose reacted
x
Produced
28
22.4
(c)
C(s)

1
22.4
(d)
14
22.4
2CO(g)
O
2x ∵ CO2  2CO
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Syllabus – cum – Comp./XI/2013–14
1–x
At end
2x
Total volume = 1 – x + 2x = 1.5 Given in Q’n
 1 + x = 1.5
x = .5 L
 Vol. of CO = 2  .5 = 1L
Moles of CO =
Q12.
1
22.4
Number of electrons present in 3.6 mg of NH 4+ are:
(a) 1.2 × 1021
Sol.
(a)
(b) 1.2 × 1020
nNH  =
4
(c) 1.2 × 1022
(d) 2 × 10-3
3.6  10 3 g
= .2  10-3 moles
14  4g
1 mole of NH4+ has = 7 + 4 – 1 = 10 moles of e-1s
.2  10-3 of NH4+ has
10  2
 10  3 moles of e-1s
10
= 2  6.023  1023  10-3
= 2  6.023  1020
= 12.046  1020
= 1.2  1021
Q13.
Caffeine has a molecular weight of 194. If it contains 28.9% by mass of nitrogen, number of
atoms of nitrogen in one molecule of caffeine is:
(a) 4
Sol.
(a)
(b) 6
(c) 2
(d) 3
100 g of caffeine has 28.9 g of N
194 g of caffeine has
28.9
 198 = 57.22 g of N
100
194 g = 1 molecule of caffeine = 57.22 g of N
1 atom of N = 14 g
 57.22 g =
Q14.
57.22
 4 (No. of atoms of N)
14
Four grams of hydrocarbon (CxHy) on complete combustion gave 12 gram of CO2. What is the
empirical formula of the hydrocarbon?
(a) CH3
(b) C4H9
[EAMCET (Medical) 2005]
(c) CH
(d) C3H8
Sol.
(d)
Q15.
The ratio of oxygen atoms to hydrogen atoms in (NH 4)2SO4 is:
(a) 1 : 1
(b) 2 : 1
(c) 1 : 2
(d) 4 : 1
O 4 1
 
H 8 2
Sol.
(c)
Q16.
A certain compound has the molecular formula X 4O6. If 10.0 g of the compound contains
5.62g of X, the atomic mass of X is
(a) 62.0 amu
(b) 48.0 amu
(c) 32.0 amu
(d) 30.8 amu
S.C.O. 208 II floor, Sec-36 D Chandigarh, Ph.3012085, Mb.9872079084
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Syllabus – cum – Comp./XI/2013–14
Sol.
Q17.
An ore contains 1.34% of the mineral argentite, Ag2 S, by mass. How many gram of this ore
would have to be processed in order to obtain 1.00 g of pure solid silver, Ag?
(b) 85.7 g
(a) 74.6 g
Sol.
(c) 107.9 g
(c) 134.0g
Ag2S  2Ag
(b)
(2  108 + 32) g  2  108 g
?
1g

248 g
216 g
?
216g of Ag is in 248g of Ag2S

1g
248
= 1.148 g of Ag 2S
216
1g of Ag is in
 For 1g of Ag, Ag2S reqd. in 1.148 g
1.34 g pure Ag2S is in 100 g of ore
1.148 g pure Ag2S is in
Q18.
100
 1.148 = 85.7g of ore
1.34
One mole of magnesium nitride on the reaction with an excess of water gives
[UP SEE 08]
(a) one mole of NH3 (b) two moles of NH3 (c) one mole of HNO3 (d) two moles of HNO3
Sol.
Mg3N2 + 6H2O  2NH3 + 3Mg (OH)2
Q19.
The compound AB contains elements A and B in the ratio of 1 : 35.5 by mass. The mass of B
that will be required to react with 2g of A is
(a) 7.1 g
Q20.
(b) 3.55 g
(c) 71 g
(d) 35.5 g
X g of calcium carbonate was completely burnt in air. The weight of solid residue formed is
28g. What is the value of X in grams?
(a) 44
Sol.
(d)
(b) 200
(c) 150
[EAMCET (Engg.) 2005]
(d) 50

CaCO3  CaO + CO2
1 mole
1 mole
100 g
56 g
100
 28
56
28 g
= 50 g
Q21.
The volume of air needed for complete combustion of 1 kg carbon at STP is:
(a) 9333.33 litre
Sol.
(b) 933.33 litre
(c) 93.33 litre
(d) 1866.67 litre
Note: Above given options are not correct.
C + O2 = CO2
1 mole 1 mole
12 g
22.4 L
22.4
 1000 = 1866 L O2
1000 g
12
100 L air has 21 L O2
100
 1866 L O2 is in
 1866 air
21
S.C.O. 208 II floor, Sec-36 D Chandigarh, Ph.3012085, Mb.9872079084
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Syllabus – cum – Comp./XI/2013–14
Q22.
Ans. 8888.92 of air.
In the commercial manufacture of nitric acid, how many moles of NO 2 produce 7.33 mol HNO3
in the reaction?
3NO2(g) + H2O(l)  2HNO3(aq) + NO(g)
(b) 10.995 mol
(a) 15.562 mol
Sol.
(c) 13.056 mol
(d) 18.556 mol
3NO2  2HNO3
3 moles 2 moles
3
moles 1 mole
2
7.33 mole of HNO3 is produced from
3
 7.33 = 10.995 moles of NO2
2
(Objective Question More than one correct answer)
Q1.
Which one of the following statements is/are correct?
(a) One mole of CH4 and 17g NH3 at NTP occupies same volume
(b) One gram mole of silver equals 108/6.023 × 10 23g
(c) One gram mole of CO2 is 6.023 × 1023 times heavier than one molecule of CO2
(d) One mole Ag weighs more than that of two moles of Ca
Sol.
(a) right
Both are 1 mole = 22.4 L
(b) x
1 gm mole = 1 mole = 108g
(c) right
(d) right
1 mole Ag = 108g
2 mole of Ca = 2  40 g
(Subjective Question )
Q23.
Glucose, or blood sugar, has the molecular formula C 6H12O6. What is the empirical formula,
and what is the percent composition of glucose?
 (Comprehension
Passage-1:- The term mole first used by Ostwald in 1896 refers for the ratio of mass of a substance
in g and its molecular weight. 1 mole of a gaseous compound occupies 22.4 litre at NTP and
contains 6.023 × 1023 molecules of gas.
Q1.
Weight of 1 atom of hydrogen is:
(a) 1.66 × 10-24 amu (b) 3.32 × 10-24 g
Sol.
(d) 3.32 × 10-24 amu
6.023  1023 atom = 1 g
1 atom =
Q2.
 (c) 1.66 × 10-24 g
1
6.023  10 23
g = 1.66  10-24 g
Avogadro’s number of Rupees can be spend in years….if 10 lacs rupees per second are
spend:
S.C.O. 208 II floor, Sec-36 D Chandigarh, Ph.3012085, Mb.9872079084
[6]
Syllabus – cum – Comp./XI/2013–14
(a) 1.91 × 1010 year
Sol.
(a)
(b) 2.91 × 1010 year
Q3.
10
6
 6.023  1023 = 1.91  1010 yr
The amount of sulphur required to produce 100 mole of H2SO4 is:
(a) 3.2 × 103 g
Sol.
.317  10 7
(d) 4.91 × 1010 year
[1 yr = 365 × 24 × 60 × 60 sec]  1 sec = .317 × 10–7 yr
106 rupees in .317  10-7 yr
6.023  1023 in =
(c) 3.91 × 1010 year
(b) 32.65 g
S + O2  SO2
(1)
SO2 + 1/2O2  SO3
(2)
SO3 + H2O  H2SO4
(3)
(c) 32 g
(d) 3.2 g
S + 3/2O2 + H2O  H2SO4 (a)
1 mole
100 mole
1 mole
100 mole
100  32 = 3200 g
Q5.
A substance contains 3.4% sulphur. If it contains two molecules of sulphur per molecule the
minimum molecular weight of substance will be:
(a) 941
(b) 1882
(c) 470.5
(d) 1411.5
Sol.
(b)
100 g of sample has 3.4 g of S, 1 molecule has 2S atoms i.e. 64g of S

3.4 g of S in 100 g sample
64 g of S in
100
 64  1882 .35g
3.4
100
 64 = 1882.35 g
3 .4
64 g of S (∵ 1 molecule has 2 S i.e 64 g)
6.
The volume of air needed to burn 12 g carbon completely at STP is:
(a) 22.4 litre
Sol.
Q7.
(a)
(b) 112 litre
(d) 50 litre
C + O2  CO2
1 mole
1 mole
12 g
22.4 L
The maximum number of atoms present are in:
(a) 4 g He
Sol.
(c) 44.8 litre
(b) 4 g O2
(c) 4 g O3
(d) 4g H2O2
(a) 4 g of He  NA
(b)
32 g of O2  2  NA atoms
4 g of O2
(c)
2  NA  4 NA

atoms
32
4
3  16 g of O3  3  NA atoms
4 g of O3 
(d)
3  NA  4 NA

atoms
3  16
4
34 g of H2O2 has 4 NA of atoms ;
4 g of H2O2 has
4 NA  4 16

 NA
34
34
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