1. Class : XI
Time : 100 min Max. Marks : 75
INSTRUCTIONS
General Remarks:
1. The questionpaper contains 16 questions and 24 pages.Allquestions are compulsory.
2. Write your answer(s) in the space given at the end of each question. Otherwise Marks will not
be awarded.
3. Each question should be done only in the space provided for it, otherwise the solution will not be
checked.
4. Use ofCalculator, Log table and Mobile is not permitted.
5. Legibilityand clarityinanswering the questionwillbe appreciated.
6. Put a cross ( × ) on the rough work done by you.
Name ________________________________ Father's Name ____________________________
Class : __________ Batch : B.C. Roll No. ___________
Invigilator's Full Name __________________________________________________
USEFUL DATA
Atomic weights:Al = 27, Mg = 24, Cu = 63.5, Mn = 55, Cl= 35.5, O = 16, H = 1, P= 31, Ag = 108, N = 14,
Li = 7, I = 127, Cr = 52, K=39, S = 32, Na = 23, C = 12, Br = 80, Fe = 56, Ca = 40, Zn = 65.4, Pt = 195
REVIEW TEST-4
Q.No 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Marks
For Office Use ……………………………. Total Marks Obtained…………………

2. Q.1 A packet ofan aritificalsweetnercontains 45.75 mg ofsaccharin which has the structuralformula
How manymoles&molecules ofsaccharinarecontainedin45.75mgofsaccharin?How manyhydrogen
atoms? [NA = 6 × 1022] [Ans. 2.5 × 10–4, 1.5 × 1020, 7.5 × 1020] [3]
[Sol: Moles of Comp. =
183
10
75
.
45 3


= 2.5 × 10–4
Molecules of compound = 2.5 × 10–4 × NA
= 2.5 × 10–4 × 6 ×1023
= 1.5 × 1020
No. of hydrogen atoms = 1.5 × 1020 × 5 = 7.5 × 1020 ]
Q.2 The subshellthat arises after f sub-shellis called g sub-shell.
(a) How manyg orbitals are present in the g sub-shell
(b) In what principal electronic shell(n) would the g sub-shellfirst occur and what is the totalnumber of
orbitals inthe shell. [Ans. (a) 9, (b) 5, 25] [3]
[Sol. For f subshell l = 3
 g subshell l = 4
(a)  totalg orbitals in g-subshells = 2l+1
= 2 × 4 + 1 = 9
(b) for principal shell, l = 4, l = n–1  n = 5
total no. oforbitals in shell = n2 = 52 = 25 ]
Q.3 For a real gas (mol. mass = 30) if density at critical point is 0.40 g/cm3 and its Tc =
821
10
2 5

K, then
calculate VanderWaal's constant a (inatmL2mol–2). [Ans. 1.6875 ] [3]
[Sol. Vc =
40
.
0
30
 75 cm3mol–1
 b =
3
Vc
 25 cm3mol–1  0.025 Lmol–1
 Tc =
Rb
27
a
8
821
10
2 5

=
025
.
0
0821
.
0
27
a
8



 a = 1.6875 ]
Q.4 The percentage byvolume ofC3H8 in a gaseous mixture ofC3H8, CH4 and CO is 39. When100 mlof
the mixture is burnt in excess ofO2. Find volume (in ml) ofCO2 produced.[Ans. 178 ml] [3]
[Sol. 100 mlgaseous mixture contain 39 mlC3H8
So, mixing volume of CH4 & CO = (100–39) = 61 ml
C3H8 + 5O2 — 3CO2 + 4H2O; CH4 + 2O2 —CO2 + 2H2O; CO +
2
1
O2 —CO2

3. 61 ml (CH4 & CO) will produce 61 ml CO2; C3H8 will produce = 3 × 39 = 117 ml
Total CO2 produce = 117 + 61 = 178 ml ]
Q.5 The angular momentumofanelectronina Bohr's orbit ofH-atomis 3.1652×10–34 kg-m2/sec. Calculate
thewavenumber interms ofRydberg constant (R) ofthe spectralline emitted whenanelectronfallsfrom
this level to the ground state.[Use h = 6.6 × 10–34 Js] [Ans. R 





9
8
] [3]
[Sol. AngularMomentum mvr =

2
nh
3.1652×10–34 =

2
10
6
.
6 34



n
n=3
  = R. Z2. 







 2
2
2
1
n
1
n
1
 = R. 12 





 2
2
3
1
1
1
 R 





9
8
Ans.]
Q.6 Use the Heisenberg's UncertainityPrinciple to calculate the uncertainity(inmeters) in the positionofa
honeybee weighing0.70gand travellingat a velocityof0.85m/s.Assumethe uncertainityinthe velocity
is 0.1 m/s.[Use h = 6.6 × 10–34 Js] [Ans. 7.53 × 10–31 m] [3]
[Sol: x × v =
m
h

4
x =
   3
34
10
1
.
0
7
.
0
14
.
3
)
4
(
10
6
.
6





x = 7.53 × 10–31 m ]
Q.7 IfPd v/sP (where P denotes pressureinatmand d denotes densityingm/L) isplotted for H2 gas (ideal
gas) at a particular temperature. If
atm
21
.
8
P
)
Pd
(
dP
d







= 10, then find the temperature in °C. [3]
[Ans. 40 K]
[Sol. PM = dRT
Pd = 





RT
M
P2

dP
)
Pd
(
d
=
RT
PM
2
 10 ;
T
0821
.
0
2
21
.
8
2



= 10
T = 40 K ]
Q.8 Radiationcorrespondingto the transition n= 4 to n= 2 in hydrogenatoms falls on acertain alkalimetal
(work function= 2.0 eV). Calculate maximumkinetic energy(in eV)ofthe photoelectrons. [4]
[Ans. 0.55]

4. [Sol. En = – 2
n
6
.
13
eV; E2 = – 2
2
6
.
13
; E4 = – 2
4
6
.
13
eV/atom
E = E4 – E2 = 2.55 eV
Absorbed energy= work function ofmetal+ KEmax
2.55 = 2 + KEmax
KE max = 0.55 eV Ans. ]
Q.9(a) What is the shortest wavelength of line (in nanometers) in the Lyman series of He+ spectrum
[use R = 1.097 × 10–2 nm–1]
(b) Whichisdenserat the same temperatureandpressure, dryairor air saturatedwithwater vapour?Justify
inbrief. [2+2]
[Ans. (a)
4
097
.
1
100

nm (b) dryair]
[Sol.

1
= 





 2
2
2
1
2 1
1
n
n
Z
RH

1
= 1.097 × 10–2(2)2 







1
1
1
 =
4
097
.
1
100

nm
(b) Dryair is denser; it has a large average molecular mass]
Q.10 For the reaction 2NH3(g) — N2(g)+ 3H2(g). What is the % ofNH3 converted ifthe mixture diffuses
twice as fast as that ofSO2 under similar conditions [Ans. 6.25] [5]
[Sol.
avg
SO
SO
mix
M
M
r
r 2
2
 
1
2
=
avg
M
64
;Mavg = 16
Let the initialmoles be one
2NH3  N2 + 3H2
1–2x x 3x
Mavg = moles
Total
.
wt
Total
; 16 =
x
2
1
17

; x =
32
1
; moles dissociated =
16
1
% dissociation = 100
1
)
16
1
(
 = 6.25% ]
Q.11 Urea (H2N–CO–NH2) is manufactured on large scale by passing CO2(g) throughammonia solution
followed by crystallization. CO2 for the above reaction is prepared bycombustion ofhydrocarbon. If
combustion of236.11 kg ofasaturated hydrocarbon(CnH2n+2) produces as muchCO2 asrequired for
production of1000kg urea, deduce molecular formula ofhydrocarbon.
2NH3 + CO2  NH2CONH2 + H2O [Ans. C12H26] [5]
[Sol: CnH2n+2  nCO2
POAC for C
n × moles of CnH2n+2 = 2
CO
n ..........(1)
Now, 2NH3 + CO2  NH2CONH2 + H2O
2
CO
n = nurea

5. 60
10
1000
2
14
10
11
.
236 3
3





n
n
Onsolving
 n = 12
C12H26 ]
Q.12(a)Arrange the correct order ofincreasing wavenumber ofthe following radiations
I.R., U.V., radiowaves, X-rays and visible light
(b) Matchthe column
Column I Column II
(A) Lyman series (P)Visible region
(B) Humphreyseries (Q)Ultraviolet region
(C) Paschen series (R) Infrared region
(D) Balmer series (S) Far infrared region
(c) Two flasksAand B ofequalvolumes maintained at temperature 300 K and 600 Kcontain equalmass
ofH2 and CH4 respectively. Calculate the ratio oftotaltranslationalkinetic energyofgas in flaskAto
that offlask B. [2+2+2]
[Ans. 4 ]
[Sol.(a)X-rays > U.V. > visible light > I.R. > radiowaves
(b) AQ, BS, CR, DP
(c) (KE)A =
2
3
A
H T
n 2
= 300
2
2
3






(KE)A =
2
3
B
CH T
n 4
= 600
16
2
3





 
 
 B
A
KE
KE
 =
600
300
2
16
 = 4 Ans. ]
Q.13 2.5 g ofa sample containing Na2CO3, NaHCO3 andsome non-volatile impurityongentle heating loses
12.4 % ofits weight. Residue isdissolved in100 mlwater and its 10 mlportion required 15 mlof0.1 M
aqueous solution ofBaCl2 for completeprecipitation ofcarbonates. Determine mass % ofNa2CO3 in
theoriginalsample.
2NaHCO3  Na2CO3 + CO2 + H2O
Na2CO3 + BaCl2  2NaCl + BaCO3  [6]
[Ans. 42.4% , Na2CO3]
[Sol: Let 3
2CO
Na
n = a & 3
NaHCO
n = b
2NaHCO3  Na2CO3 + CO2 + H2O
 moles of CO2 produced =
2
b
moles of H2O produced =
2
b
total loss in weight = 





 44
2
b
+ 





18
2
b
= 31 b
 5
.
2
100
4
.
12
 = 31b

6. b = 0.01
Now in residue moles of Na2CO3 = a +
2
b
= a + 0.005
Na2CO3 + BaCl2  2NaCl + BaCO3 
moles of Na2CO3 = moles of BaCO3
a + 0.005 = 15 × 0.1 ×
10
100
× 10–3
a = 0.01
 wt. of Na2CO3 = 0.01 × 106 = 1.06 gm
% Na2CO3 = 100
5
.
2
06
.
1
 = 42.4% Ans. ]
Q.14(a) At 273 K and 8 atm pressure, the comprassibility factor for a gas is 0.8. Calculate the volume of
1 millimolesofgas at thistemperature and pressure.
(b) Fromhis exceptional'mathematicalskills', Mr. Gupta was able to obtain the following expression for
 (the amplitude ofelectronwave) bysolvingSchrodinger equation. However he wantsto know those
'r'whereprobabilityoffinding theelectronis zero. Sincehe is'exhausted'inhiswork, helphimto find the
values of'r' acting as nodes( 2 = zero).
  2
/
2
2
/
3
0
e
)
12
8
)(
1
(
a
Z
4
16
1 

















where a0 & Z are the constant inwhich answer can be expressed &
0
a
Zr
2

 [3+5]
[Sol: (a) PV= ZnRT
8 × V = 0.8 (10–3) (0.0821× 273)
V = 2.24 ml Ans.
(b)  =   
 
12
8
1
4
16
1 2
2
/
3
0














a
Z
e–/2
At node, = 0
   
 
2
2
/
2
2
/
3
0
12
8
1
4
16
1


















 


 e
a
Z
= 0
 ( – 1) (2 – 8+ 12) = 0
 ( – 1) (2 – 8+ 16 – 4) = 0
 ( – 1) [( – 4)2 – 4] = 0
 ( – 1) [( – 4)2 – 22) = 0
 ( – 1) ( – 4+2) ( – 4 – 2) = 0
 ( – 1) ( – 2) ( – 6) = 0
so  = 1 or  = 2 or  = 6

0
2
na
Zr
= 1 or
0
2
na
Zr
= 2 or
0
2
na
Zr
= 6
 r =
Z
na
2
0
or r =
Z
na0
or r =
Z
na0
3
Ans.]

7. Q.15(a) Calculate the ratio oftime periods in first and third orbits ofhydrogen atom.
(b) The electroninthefirst excited state (n1 =2) ofH-atomabsorbs a photonand is furtherexcited (n2).The
Debroglie wavelength ofthe electron in this excited state is 1340 pm. Find the value ofn2.
[3+5]
[Ans.(a) 1/27, (b) 4]
[Sol(a)
3
3
1
3
1
n
n
T
T








 =
27
1
3
1
3







(b) rn
nr
n
1
2 r



   
53
14
.
3
2
1340
 ]
Q.16 The apparatus shown consists of three temperature jacketed 1 litre bulbs connected by stop cocks.
BulbAcontains a mixture ofH2O(g), CO2(g) and N2(g) at 27°Cand a totalpressure of547.2 mmHg.
Bulb B is empty and is held at a temperature –23°C. Bulb C is also emptyand is held at a temperature
of–173°C. The stopcocks are closed and the volumes oflines connecting the bulbs is zero.
Given: CO2(g) converted into CO2(s) at –78°, N2(g) converted into N2(s) at –196°C & H2O(g)
converted into H2O(s) at 0°C.
[Use R = 0.08 atm-litre/mole·K]
(a) The stopcock betweenA&B is opened and the systemisallowed to come to equilibrium. The pressure
inA& B is now 228 mmHg. What do bulbsA& B contain?
(b) How manymoles ofH2O are in system?
(c) Bothstopcocksare opened and thesystemis againallowedto equilibrium. The pressurethroughout the
systemis 45.6 mmHg. What do bulbsA, B and C contain?
(d) How manymoles ofN2 are in the system? [8]
[Sol. Total initial no. of moles in bulbA= ntotal =
300
08
.
0
1
760
5
.
547








=
300
08
.
0
1
72
.
0


= 0.03
After openingthe cock:
In bulbA
)
CO
N
( 2
2
n  =
24
1
760
228







= 0.0125
I bulb B )
CO
N
( 2
2
n  =
250
08
.
0
1
760
228








=
250
08
.
0
3
.
0

= 0.015
O
H2
n = 0.03 – (0.0125 + 0.015) = 0.0025

8. After opening C  2
N
n =
300
08
.
0
1
760
6
.
45


= 0.0025
2
N
n in B 2
N
n =
250
08
.
0
1
760
6
.
45


= 0.003
2
N
n in C 2
N
n =
100
08
.
0
1
760
6
.
45


=
100
08
.
0
06
.
0

= 0.0075 ]