Class : XIII (Sterling)
Time : 3 hour Max. Marks : 60
INSTRUCTIONS
1. The question paper contain pages and 3-parts. Part-A contains 6 objective question , Part-B contains 2
questions of "Match theColumn" type and Part-C contains 4 subjective type questions.All questions are
compulsory. Please ensure that the Question Paper you have received contains all the QUESTIONS
and Pages. If you found some mistake like missing questions or pages then contact immediately to
the Invigilator.
PART-A
(i) Q.1 to Q.6 have One or More than one is / are correct alternative(s) and carry 4 marks each.
There is NEGATIVE marking. 1 mark will be deducted for eachwrong answer.
PART-B
(iii) Q.1 to Q.2 are"Match the Column" typewhich mayhave oneor more than onematching options and
carry8 marks foreach question. 2 marks will be awarded for each correct match within a question.
ThereisNONEGATIVEmarking.Markswillbeawardedonlyifallthecorrectalternativesareselected.
PART-C
(iv) Q.1 to Q.4 are "Subjective" questions. There is NO NEGATIVE marking. Marks will be awarded
onlyifall the correct bubbles are filled in yourOMR sheet.
2. Indicate the correctanswer for each question byfillingappropriate bubble(s) inyour answer sheet.
3. Use onlyHB pencil fordarkeningthe bubble(s).
4. Use ofCalculator, LogTable, SlideRule and Mobile is not allowed.
5. Theanswer(s)ofthequestionsmustbemarked byshadingthecirclesagainst thequestionbydarkHBpencil
only.
PART TEST-1
PART-B
For example if Correct
match for (A) is P, Q; for
(B) is P, R; for (C) is P
and for (D) is S then the
correct method for filling
thebubbleis
P Q R S
(A)
(B)
(C)
(D)
PART-C
Ensure that all columns
(4 before decimal and 2
after decimal) are filled.
Answer having blank
column willbe treated as
incorrect. Insert leading
zero(s) if required after
rounding the result to 2
decimalplaces.
e.g. 86 should befilled as
0086.00
.
.
.
.
.
.
.
.
.
.
PART-A
For example if only 'B'
choice is correct then,
the correct method for
fillingthebubbleis
A B C D
Forexampleifonly'B&
D' choices are correct
then, the correct method
forfillingthebubblesis
A B C D
The answer of the
question in any other
manner (such as putting
, cross , or partial
shading etc.) will be
treated as wrong.

Class - XIII Chemistry Part Test - 1
PART A
Q.1 Calculatetotalmaximummass(kg)whichcanbeliftedby10identicalballoon(eachhavingvolume82.1
lit. and mass of balloon & gas = 3 kg) at a height 83.14 m at Mars where g = 5 m/s2 & atmosphere
containsonlyAr(At. wt.40).AtMars temperature is 10K and densityofatmosphereat groundlevel is
821
.
0
2
k gm/lit. [Given : e–0.1 = 0.9] (Assume dH = d0
RT
Mgh
e
to be applicable).
(A*) 2000 × 0.81–30 (B) 1970 (C) 2000 × 0.9 –30 (D) 2000 × 0.81 – 3
[Sol: (A) d = d0
RT
Mgh
e

d = 821
.
0
2 10
314
.
8
14
.
83
5
10
40 3






e
= 821
.
0
2
e–0.2 =
2
)
9
.
0
(
821
.
0
2

Now at height h = 83.14 m
V × d = load
of
mass
filled
gas
of
mass
balloons
of
mass 







 






 


2
)
9
.
0
(
821
.
0
10
1
.
82


= (3 × 10) + m
 m = (2000 × 0.81) – 30 Kg ]
Q.2 How manymg of quick lime is required to remove hardness of 1 kg of hard water having 366 ppm of
HCO3
– and contains Ca2+ as the only cation
(A) 72 mg (B) 84 mg (C*) 168 mg (D) 170 mg
[Sol: In 1 Kg water
–
3
HCO
w = 366 × 10–3 gm
–
3
HCO
n = 6 × 10–3
CaO + Ca(HCO3
–)2  2CaCO3 + H2O
 moles of CaO required =
2
n –
3
HCO
= 3 × 10–3
 wt. of CaO required = 3 × 10–3 × 56
= 168 × 10–3 gm
= 168 mg [C] ]
Q.3 A 10 litre box contains O3 and O2 at equilibrium at 2000K. Kp = 4.17 × 1014 for 2O3 3O2.
Assume that 3
2 O
O P
P  and if total pressureis 7.33 atm, then partial pressure of O3 will be
(A) 9.71 × 10–5 atm (B*) 9.71 × 10–7 atm
(C) 9.71 × 10–6 atm (D) 9.71 × 10–2 atm
[Sol: 2O3 3O2
Kp = 2
3
3
2
O
O
P
P
Here 3
2 O
O P
P  so the total pressure 2
O
P


 4.17 × 1014 =
 
2
3
O
3
P
33
.
7
3
O
P = 9.71 × 10–7 atm ]
Q.4 

 
 KOH
Br2
If the reactant is (d) (dextrorotatory) then in final product 
(A)inversionofconfigurationtakes place
(B)racemisationtakes place
(C*) retentionofconfigurationtakes place
(D) none of the above
[Sol: r.d.s.is1,2shift,shiftingtakes placewith retentionofconfiguration ]
Q.5 Select the correct statement(s)
If S = f (E, V, n1, n2) &A= f (T, V, n1, n2)
(A*) S is equal to f (E, V, n1, n2) (B) S is equal to f (E, V, n1, n2)
(C*) Ais equal to f (T, V, n1, n2) (D) Ais equal to f (E, V, n1, n2)
[Sol: S = f (E, V, n1, n2)
& E, V, n1, n2 all are extensive vairable
 S = f (E, V, n1, n2)
A= f (T, V, n1, n2)
hereTisintensivevariable
 A= f (T, V, n1, n2) [A, C] ]
Q.6
Compound Bhas a neutralisationequivalent112. G is adichloro alkane.
(A*) A = , B = , C = , E = ; F =
& G = ClCH2CH2CH2CH2Cl
(B) A = , B = , C = , E =
& G = ClCH2CH2CH2CH2CH2Cl
(C) A = , B = , C = , E = , F =
& G = ClCH2CH2CH2CH2CH2Cl

(D) A = , B = , C = ;
E = ; F = & G = ClCH2CH2CH2CH2CH2Cl
[Sol:
PART B
Q.1 Matchthefollowing
ColumnI ColumnII
(A) Canaryyellowprecipitate with (NH4)2MoO4 (P) NO3
–
(B) Brownringtest (Q) NO2
–
(C)Acidradical whichevolves gas with conc. HCl (R) AsO4
3–
(D)Acidradical whichgives gas withdil. H2SO4 (S) PO4
3–
[Ans: (A)  R, S; (B)  P, Q; (C)  P; (D)  Q]
[Sol: (A) (NH4)2MoO4 + AsO4
3– + H+  (NH4)3Mo12AsO40
Canaryyellow ppt.
(NH4)2MoO4 + PO4
3– + H+  (NH4)3M12PO40 
Canaryyellow ppt.
(B) NO3
– + Fe2+ + H2SO4  Fe3+ + NO + SO4
2– + H2O
Fe2+ + NO [Fe(NO)]2+
Brownring

NO2
– + CH3COOH  HNO2 + CH3COO–
HNO2  NO + HNO3 + H2O
Fe2+ + SO4
2– + NO [Fe(NO)]2+ Brown ring
(C) NO3
– + H+ NO2 + H2O
(D) NO2
– + H+  HNO2  NO + HNO3 + H2O ]
Q.2 Matchthefollowing
List I List II
(A) 
 
CuCl (P)Freeradical mechanism
(B) CH3–CH=CH2 (Q) Non Classical Carbocation
 Cl2/
Cl
|
CH
CH
CH 2
2 

(C) CH3–CH=CH2 


 
 Cu
/
Zn
/
I
CH 2
2 (R) Carbenoid
(D) CH3–CH=CH2 





 

OH
/
NaBH
)
ii
(
O
H
/
)
OAc
(
Hg
)
i
(
4
2
2
(S)Through carbocation
[Ans: (A)  P; (B)  P; (C)  R; (D)  Q]
[Sol: (A) PhCl
CuCl
N
Ph 2 


Mechanism
2
2
o
2 CuCl
N
Ph
CuCl
Cl
N
Ph 





CuCl
PhCl
CuCl
Ph 2
o



(B) Cl2 
 2Clo
(C)

(D)
PART C
Q.1 1.0gofamixturecontainingSb2O3 andSb2O5 andsomeinert impurities required 100ml0.01 Niodine
fortitration.Theresultingsolutionis thenacidifiedandexcessofKIwasadded.Theliberated I2 required
100 ml 0.02 M Na2S2O3·5H2O for complete reaction. Calculate the % of Sb2O5 in the mixture. The
reactions are [at. wt. Sb = 122]
Sb2O3 + 2I2 + 2H2O  Sb2O5 + 4H+ + 4I–
Sb2O5 + 4H+ + 4I–  Sb2O3 + 2I2 + 2H2O
[Sol: Meq of I2 used = 100 × 0.01 = 1 meq
Let meq of Sb2O3 and Sb2O5 are a and b, on addition of I2 to mixture Sb+3 converts into Sb+5
Meq. of Sb+5 formed = meq of I2 = 1 meq = meq of Sb2O3
a = 1 meq
After reaction,.mixturecontains all theSb in+5oxidation state.
Meq of Sb+5 = meq of liberated I2 = meq of hypo solution used
a + b = 100 × 0.02 × 1 = 2 meq
b = 1 meq =
4
324
.
0
4
324
1000
1

 g = 0.081 g
 Sb2O5 = 8.1 % ]
Q.2 Aballooncontaining1moleairat 1 atminitiallyis filled furtherwithairtill pressure increasesto 3 atm.
The initial diameter of the balloonis 1 m and the pressure at eachstate is proportion to diameter of the
balloon.Ifballoonwillburst whenpressureincreasesto7 atm.Calculatethenumberofmolesofairthat
must be addedafterinitial condition toburst theballoon.
[Sol: P diameter
d = Kp
Initially 1 = K × 1
K =
atm
m
1
For Pf = 7 atm

d = 7m
V =
3
2
7
3
4






 m3
7 ×
3
2
7
3
4






 = nf RT
T ..................(3)
Dividing (3) by (1)
i
f
n
n
= 7 × 7 × 7 × 7
nf = 2401
 moles added = 2400 ]
Q.3 Thekeyreactioninthemanufactureofsyntheticcryolightforaluminiumelectrolysisis
HF(g) + Al(OH)3 (s) + NaOH (aq)  Na3AlF6(aq) + H2O(l)
Assuminga96%yieldofdried,crystallized product, whatmass (inkg)ofcryolitecanbeobtainedfrom
the reaction of 351 kg ofAl(OH)3, 1.10 m3 of 50.0% by mass aqueous NaOH (d = 1.50 g/mL), and
225 m3 of gaseous HF at 312.08 kPa and 87oC? (assume that the ideal gas law holds)
[Given: Al = 27, O = 16, H = 1, Na = 23, F = 19, R = 8.3 JK–1 mol–1]
[Sol: 6HF + Al(OH)3 + 3NaOH  Na3AlF6 + 6 H2O(l)
moles ofAl(OH)3 =
78
10
351 3

= 4500
moles of NaOH =
 
40
5
.
0
5
.
1
10
1
.
1 6



= 20625
moles of HF =
RT
PV
=
360
3
.
8
225
312080


= 23500
HereHFislimitingreagent
so moles of Na3AlF6 formed =
100
96
6
23500
 = 3760
wt. of Na3AlF6 formed =
1000
210
3760
= 789.6 kg ]
Q.4 The vapour pressure of water at 300 K is 25 torr. If the standard state pressure is defined as 1 bar (750
torr) estimate the G° [inkJ/mol] for the process.
H2O (l)  H2O (g) at 300 K
(Neglect variation of H and S with pressure for liquid)
Use [R = 8.314 JK–1 mol–1 ; X
og
X
n
l
l
= 2.3, log 3 = 0.48] [Ans. 8.49 ]
[Sol: G0 = –2.3RT log Kp
= – 2.3 × 8.314 × 300 log O
H2
P (bar)
= – 2.3 × 8.314 × 300 log 





750
25
= – 2.3 × 8.314 × 300 log
30
1
= 8490J/mol = 8.49 kJ/mol ]