kvpy sol-20101

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CLASS-XII_STREAM-SB+2_PAGE # 1

PART-I (1 Mark)MATHEMATICS

1. I = !"

#

$

&

10

01

A = !"

#$&

0i

i0

A2 = !"

#$&

10

01

A3 = !

"

#$&

0i

i0

A4 = ' ( A4n = '

))))' + A + A2 + A3 = 0

=)' + A + A2 + A3 + ............+ A2010

= (' + A + A2 + A3) + A4 (' )+ A + A2 + A3) ..... + A2008 (' )+ A + A2)

= 0 + !"

#$&

0i

i0= !

"

#$&

0i

i0.

ANSWER KEYHINTS & SOLUTIONS (YEAR-2010)

Ques. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Ans. C B D B C B B A C B B C A C D

Ques. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30Ans. D D D C A A A B C D A D A B D

Ques. 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45

Ans. D B C B B A A D A C B C D A B

Ques. 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60

Ans. B D A A A C B C D B A B A C A

Ques. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75

Ans. A A C A C C A A C D A A C B B

Ques. 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90

Ans. D C A A D C B B C B C A B D C

Ques. 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105Ans. D A B B A A B D B C A A D D

Ques. 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120

Ans. C B A B D B C D B B D A C A A

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2. Let a, ar, ar2 are sides.

Case 1 : r > 1

a + ar > ar2

( r2 r 1 < 0

( r * ++

,

-

.

.

/

01

2,

2

51

Case 2 : r = 1 equilateral triangle

Case 3 : r < 1

ar2 + ar > a

( r2 + r 1 > 0 ( r * ) ++

,

-

.

.

/

0 23

2

51,

2

51

4 )r lies in the interval

!

!

"

#

.

.

/

0 22

2

52,

2

51.

3. Number of diagonals passing through centre = 6

4 Number of rectangles = 6C2= 15

4. 25 2, 3 + 45 , 3 + 45 2, 5 5 5 2

6 1 3 i , 1 + 32 i , 1 32 i, 6

where 5 =2

i31 2, 5 2 =

2

i31

1 + 32 i , 1 32 i are conjugate of each other..

4 Least possible degree = 5

5. Tangent at (2, 1)

y.2 = 2 (x + 1)

( y = x + 1

( x y + 1 = 0

equation to circle

(x 1)2 + (y 2)2 + 7(x y + 1) = 0

putting x = 14 + (y 1)2 + 7( y) = 0

( y2 4y 7y + 8 = 0

D = 0 ( (7)+ 4)2 = 32

y =4

D)4( 827

4 y =2

024 28 = 2 2

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6. 9BKN = 6

9

1

APB

BKN:

9

9 +

,

-./

0:4:

3

1

AB

BN

1

2

BN

AN

( 9 APB =1

9 6

=1

54

4 9 ABC = 108.

7. 2

2

2

2

b

y

a

x2 = 1 , a > b > 0

h =3

x, k =

3

y

2

2

2

2

b

)k3(

a

)h3(2 = 1 which is ellipse

8. 1

6

1

7 sin1cos;

9. (1 + tan 1) (1+ tan 2) .........(1+ tan 45)

(1 + tan A) (1 + tan B) = 2 if A + B = 45

Ans = 223

10. f is differentiable function .

f?(a) f?(b) > 0 ( f is increasing at x = a and b

or f is decreasing at x = a and b

minimum number of roots of f?(a) = 0 in (a, b) is 2,

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11. When x is less than 49 then f(x) has negative value.

Which is not possible.

when x > 49

then f?(x) = 49 (x 41)48 + 41 (x 49)40

+ 2009 (x 2009)2008

so sing of f?(x) does not change

f?(x) > 0( non real except one positive root.

12.

f?(x) is given smooth curve hence differentiable i- given domain

By first derivative test

x = a, c is point of local maxima

x = b is point of local minima

13. @ :3

i

4dx)x(f

shaded area = Trapezium Area Area under curve

= 2

)13(2

7

2

5+,

-./

02

4 = 2

14. 'n

= dx.)x(log

e

1

n

@

'n

= dxx.x

)x(logn)x(logx

e

i

1ne

1

n

@

'n

= (e 0) n 'n

1

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'n

+ n 'n1

= e

'2011

+ 2011 '2010

= e

4 '100

+ 100 '99

= e

15. x2 + y2< 100 ( Area bounded = 100 >

sin (x + y) > 0Hence common area of region bounded = 50 > as in half the region sin (x + y) > 0 and in other half

sin (x + y) < 0

16. n(S) = 7C3

= 35 n (* ) = 7 + 7 + 7 = 21 P (* ) =5

3

35

21:

17. w).vu(

A , vu

A =230

112

kji

= )6(k)4(j)32(i 22 = 53

greatest when x2 + y2 = 1

fmax.

= 17 Ans. (D)

18.

Number of ways = 4P2 4P

2 10 = 1440

19. 1 + x2 + x4 + .........+ x2010

=2

10062

x1

)x(1

= )x1)(x1(

)x(1 21006

2

=)x1(

)x1(

)x1(

)x1( 10061006

2

2

= (1 + x1006))x1(

)x1(

)x1(

)x1( 503503

2

2

= (1 + x1006) (1 + x + x2 + ...............+ x502) (1 x + x2 x3 + ............ + x502)

4 n 1 = 502 n = 503

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20. an

= 3an1

+ 1

= 3(3an2

+ 1) + 1

= 32 an2

+ 3 + 1

= 32 (3an3

+ 1) + 3 + 1

= 33 an3

+ 32 + 3 + 1

aN

= 3n a0

+ 3N1 + 3N2 + .... 1

Method - Ia

2010= (1 + 3 + 32 + 33 + 34) + ....... + 32009

= (1 + 3 + 32 + 33 + 34)+ 35 (1 + 3 + 32 + 33 + 34)+ 310 (1 + 3 + 32 + 34 + 35) +.....+32005 (1+ 3 + 32 + 33 + 34)

each (1 + 3 + 32 + 33 + 34) is divisble by

Method-II

a2010

=2

1)1242(

2

1)243(

2

13

2

13 40240240252010 2:::

A

= 11 k (k * I)

Hence remainder is 0

PHYSICS

21.

F B1mg B

2(m + M)g = M.

m

mg1

F = B1mg + B

2(M + m)g + B

1Mg

= B1g(M + m) + B

2g(m + M)

= (B1g + B

2g) (M + m)

22. T =k

1

mm

mm2

21

21++,

-../

0

2>

we can use concept of reduced mass.

23. 2T cos= = ma ;2

2

x4

xmT2

2

3

= a

If x < < < . f =

mT4

21>

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24. AU C = mean distance between Sun and Earth

and Time period = 125 year

T2D r3

(1 yr)2D (1AU)3

(125)2

D

3

a

2

4.0r

!"

#

$%

& 2

4 25 = !"

#$%

& 2

2

4.0ra

ra

= 50 0.4

= 49.6.

25. Zero

26. 3R

KQrE D r inside

2r

KQE D 2r

1outside.

27. There is no net effect of outside charge.

29. tanC =h

r

from snell's law

sinC =B

1

r =1

h

23

30. B2

> B3

B1

sin i1

= B2

sin r

B1

< B2

from diagram2

3

2

1

B

BE

B

B

For condition of TIR

B2

> B3

and B1> B

3

So, B2

> B1

> B3

.

31. A = T/t0

2

AA = 42

120000=

2222

10100120

A

A= 7500 (D).

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32. 5 2 A = g

A

g:5

f =>2

1

A

g

= 2105.2

10

2

13

A>= 3.18

33. dQ = 0

dU = dw

++

,

-

.

.

/

023 dV

V

anCdT

2

2

= dVV

an

nbV

nRT2

2

!!"

#

$$%

&3

3

CdT =nbV

nRT3

.dV

@ @ 3:3 nbVdV

T

dT

nR

C

nTnR

C3 = n(V nb) + C

n (V nb) nR

CnT = K

[n(V nb) + n TC/nR] = K

ln(V nb) TC/nR = K

(V nb) TC/nR = constant Answer is (C) option.

35.

v0

= 39.6 kmph = 39.6 185 = 2.2 5 = 11 m/s

t1

=330

d

t2

=330

3011d A3=

330

d

330

3011A

9 t = t1 t

2= 1

time interval = 30 1 = 29 second

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36. v =M

RT

4

1

32

2

M

M

v

v

0

H

H

0 :::

37. d = 0.1 mm

D = 1m

7 = 600 nm

given IR

= 75% of maximum = 75% of 4' = 3'0

'R

= 3' = ' + ' + 2 ' cos H

H =3

>( 9x =

D

dy=

>

7

2

3

>=

6

7

y =d6

D7= 3

9

101.06

1060013

3

A

A( y = 1 mm

38.

vcom

=21

01

mm

0vm

2

2( x

i=

21

21

mm

)(m)0(m

2

2

xcm

= xi+ v

comt

xcm

=21

01

21

2

mm

tvm

mm

)(m

22

2

39. According to given information

tan = =r

a

r = a cot

A B Camv

rr

mv2= qvB ( )v =

m

qBr( )v =

m

cotqBa =

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CHEMISTRY

41. [Co(dien)Cl3]

dien = NH2 CH

2 CH

2 NH CH

2 CH

2 NH

2

[Co(dien)Cl3] have two geometrical isomers cis and trans.

cis form is optically active but trans form is optically inactive.

42. The ligands which allow back bonding to sufficient extent are called ! acid acceptors.Example CO, CN, NO etc.

43. O22 : Total electron = 18 ; Bond order = 1.

Peroxide (O22) : (I 1s)2 (I *1s)2 (I 2s)2 (I *2s)2 (I 2p

z)2 (>2p2

x= >2p2

y) (>*2p

x2 = >*2p2

y)

Bond order = !"

#$%

&

2

NN AB=

2

810= 1.

44. E =7

hc=

1

1031062.6 834 A3= 1.988 1025 J.

45. t1/2

(half-life) is independent of initial concentration so reaction is first order.

46. ClF3

F

Cl

F

87.5

87.5

F

F

..

..

F nearly 'T' shaped.

47. 3AlCl

O||

ClCR

J K

Friedal craft acylation

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48. > > >

Order of stability of conjugate bases

> > >

49. anti conformation is most stable

50. ePaTh0

1

234

91

234

902K

51. Entropy change is positive, but enthalpy does not change.

52. On increasing temperature, change in activation energy is not significant.

53. For FCC unit cell number of atoms per unit cell

Z = 8[corner] 8

1+ 6 [face center]

2

1= 4.

For body center unit cell number of atoms per unit cell

Z = 8[corner] 8

1+ 1 [body center] 1 = 2.

54. CO2

CO +2

1O

2;

1CK = 9.1 1012 .....(i)

H2O H

2+

2

1O

2;

2CK = 7.1 1012 .....(ii)

Target eq.

CO2

+ H2

CO + H2O K

C= ?

Target eq. can be deduce by

[Eq. (i) Eq(ii)] ( KC

=2

1

C

C

K

K

=12

12

101.7

101.93

3

A

A= 1.28.

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55. R J K P

t=0 R0

0

t=t (R0 x) x

Rate = dt

]R[d= K[R]1

=dt

dx= K[R

0 x] = @

x

00

dx.)xR(

1= @

t

0

dt.K

= ln ++

,

-

.

.

/

0

xR

R

0

0= Kt

[R] = (R0 x) = R

0eKt

56. PCl3F

2:

57. (I) & (IV) are enantiomeric pair

I IV

58. 2NO2(g) J K 2NO (g) + O

2(g)

NO2is reactant so its concentration is decrease with time while NO and O

2are product so their concentration

increases with time.

Formation of moles of NO is double than O2.So X = NO, Y = O

2and Z = NO

2.

59. Cyclic

Planar

Complete conjugation

(4n + 2) >e

Tropylium carbocation

60. Anti addition takes place.

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PART-II (2 Mark)MATHEMATICS

81.

n

4/1

2/1

x2

1x +

,

-./

02 =

n

4/3

2/n

x2

11x +

,

-./

02

= xn/2LM

LN

O

LP

LQ

R+,

-./

022+

,

-./

02+

,

-./

02 .........

x2

1c....

x2

1c

x2

1cc

4/3rn

2

4/32n

4/31n

0n

nc0

, 22

n1

n

2

c,

2

care in AP

( 1,2

n,

8

)1n(nare in APAP

( n = 1 +8

)1n(n

( 8(n 1) n (n 1) = 0

( (n 1) (8 n) = 0 ( n = 1 or n = 8

but n = 1 i.e. not posible

( n = 8

Now

8

4/1

2/1

x2

1x +

,

-./

02 =

r

4/1

8

0r

r82/1r

8

x2

1)x(c +

,

-./

0S

:

=4

r34

r8

0r

r8 x

2

1c +

,

-./

0S

:

required number of terms = 3 (r = 0, 4, 8)

82. Roots of 2x2 + 2x = 1 = 0 are ,2

i183

It will satisfies the (x + 1)n r = 0

(

n

12

i1+,

-./

02

8= r

(

n

12

i1+,

-./

02

8= r, r * R

(

n

4

in

e2

1

++

,

-

.

.

/

0

++,

-../

0>

8

= r

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n

4

i

e++

,

-

.

.

/

0 >8

will be real only if n is multiple of 4.

( For n = 4000

( r =

4000

2

1

++,

-

../

0

= 10004

1

83. y = ax + b and 1b

y

a

x 22:2

fig. 1 and 3 incorrect ()a < 0 b the equation of line y = ax + b)

Now for a > 0 and b < 0 fig. 2 is correct

84. Area of cyclic quadratic and is maximum only if area of 9ABC and area of 9ADC is maximum

( Area of 9ABC is maximum if ABC is equilateral 9 .

( AC = 2R sin 60 = 3 R.

maximum Area of 9 )ABC =2)R3(

4

3=

2R4

33

maximum area of 9 ADC =2

1(AC) (DM)

=2

1++

,

-

.

.

/

060cot

2

R3)R3(

=4

R3 2

area of ABCD = 222 R3R4

3R

4

33:2

85.

f(x) is monotonic increasing in [a, T )

( f?(x) > 0 U x * [a,T )

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( f?(x) is monotonic decreasing in (T , b] ( f?(x) < 0 U x * (T , b]

Now g?(x) =x

)x(f

g?(x) = 2x

)x(f)x(xf?

h(x) = xf?(x) f(x)

( h?(x) = f?(x) + xfV (x) f?(x)

( h?(x) = xfV < 0 { x > 0 and fV (x) < 0 concavity}

( h(x) is M.D.

So, at x = a it may be M.I. followed by M.D. till x = b

or

M.D. through out a

But cant be M.I. followed by M.D.

So (B) is correct answer.

86.

Let radius of base of given cone is r and length

as h ( V1

=3

1>r2h ........(1)

Let radius of base 1 and R1 and height H. For the cane with apex O

V2 = 31 >R2H ..........(2)

NowHh

h

R

r: ( i

r

R

h

H:

( H = hMNO

PQR

r

R1

V = +,

-./

0>

r

R1hR

3

1 2

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++

,

-

.

.

/

0>:

r

R2R2h

3

1

dR

dv 2

= 0 ( R =3

r2

For maximum volume

R 3

r2( V

2= +

,

-.

/

0>

r

R1hR

3

1 2

= +,

-./

0>

3

21

9

r4h

3

1 2

= h81

4 2> .......(2)\ (27

4

1V

V2 :

87. f?(x) = 1 + f(x) (dx

dy= 1 + y

( n (1 + y) = x + c

( y = ex + c 1

( y = 7)ex 1

( f(x) = 7)ex 1 ..... (1)

Now f(x) = x + @x

0

dt)t(f

7ex 1 = x + 7et (t)0x

7ex 1 = x + 7ex x 7

( 7)= 1

( f(x) = ex 1

for f(x) = 0 ( ex 1 = 0

( W x = 0

88.

A =2

1.>)X

2

>+

2

1.>)X

2

>): )

2

2>

89. Let P is origin and p.v. of A, B, C are a, b, c respectively

Given 0PC3PB2PA

:22K

( 0c3b2a :22

...... (i)

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Area of 9ABC =2

1|a

b

+ b

c

+ c

a

|

=2

1|a

++,

-../

0

2

ac3

+ ++,

-../

0 3

2

ac3

c

+ c

a

|

=2

1|2

3( c

a

) 0 0 +2

1( c

a

) + c

a

|

=2

3|c

a

| ...... (ii)

Area of 9 APC =2

1|c

a

|

(APC

ABC

9

9=

1

3

90. 6m + 2m+n. 3m + 2n = 332 = 4 83

( 2m2. 3m + 2m+n2 3m + 2n2 = 83 .......(1)

83 is prime number so that this is possible only if

m = 2

( by (1) 32 + 2n 32 + 2n2 = 83

( 2n. 9 + 2n2 = 74

( 2n2 (36 + 1) = 74

( 2n2 = 21

( n = 3Hence. m2 + mn + n2

= 4 + 6 + 9 = 19

PHYSICS

91. h10upto........g

Vr

g

Vr

g

V th20

420

220 3

!!"

#

$$%

&222

=g

V20 [1 + r2 + r4 + .............upto 10th] h

= hr1

)r(1h2

2

102

3!!"

#

$$%

&

3

3

= hr1

r1h2

2

20

3!!"

#

$$%

&

3

3.

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92. 2r

GMm=

r

mv2( v =

r

GM

T =V

r2>T =

GM

r2

3

>

T = rGM

r2>

= GM

r

2

3

>

geff

= g 5 2 Re cos2H

f(g 5 2 Re cos260) = (g 5 2Re cos20)

fg g = 5 2R !"

#$%

&3 1

4

f

g[f 1] =4

R25[f 4] ( 2R

Gm(f 1) =

4

R25(f 4)

)4f(

)1f(GM4

2 35

3

= R3

( )4f(

)1f(GMT

2

2

3>

3

= R3

)4f(

)1f(GMT2

2

3>

3= R3 Y =

3R3

4

M

>=

)1f(GT4

)4f(32 3

3>

93. a = 2r 2

32

2

q

Kq2

2

q

Kq2cos30

= r3

kr = 2

2

q

kq2

cos30

k.r.3r2 =04

2

>Z.q2

2

3

r =

3/1

0

2

k12

3q++

,

-

.

.

/

0

>Z

94. Consider the infinite ladder circuit shown below :

6

6

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Let the equivalent impedence of circuit be ZSo, Z = wL + Z?

Z = 5 L +C

C

XZ

ZX

2= 5 L +

C

1Z

C

1Z

52

5A

On solving we get,

Z =C2

LC4CLLC 222 3585

For Z to be purely inductive 0LC4CL 222 :35 = 0

5 =LC

2Ans.

95.

13

1

V1=

R

13(

R

1

v1

3: (

1

Rv1

3

B:

)VR2(V

1

1f 33

B3 =

R

1

3

B3

++,

-../

0

3

B3

B3

3

B3:

1

RR2

R

1

V

1

f

)RR2R2(

)1(

R

1

B33

33

3= !

"

#$

%

&

3

B3

3

2

1

R

1

!"

#$%

&

3

B333

)2(

)2(

R

)1(= ++

,

-../

0

3

3+,

-./

0 3

2

2

R

1( v

f=

)1(2

)2(R

3

3

96. V(x) =2

1kx2 V

0cos +

,

-./

0

a

x

G =dx

dV3

= kx + V0 sin +,

-

./

0

a

x

. a

1

Since, x

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97. 9Q = 9U + 9W

nCPdt = nC

vdt +

T3

3

1

)TT(nR 12

Heat capacity in this process

CP

= Cv+

T31

nR.

98. W = Area of 9ABC

W = +2

1(P

2P

1) (V

2V

1)

P1 BA

C

V2V1

P2

{It is clockwise}Heat released at CA :It is at constant volumeSo, Q = nC

VdT

Q = n2

3R (T

A TT

C)

Q1 = 2

3

(P1 V2

P2V2)

Heat released at A B (Constant pressure)

Q2= nC

PdT =

2

5nR (T

B TT

A)

Q2

=2

5(P

1V

1 P

1V

2)

Heat absorbed at BC9Q = du + 9 W du = 0Q

BC+ Q

1+ Q

2=

9W

QBC

= 9 W + Q1

+ Q2

=21 (P

2 P

1)(V

2V

1)

25 (P

1V

1P

1V

2)

23 (P

1V

2P

2V

2)

=2

1[(P

2P

1)(V

2V

1) 5P

1V

1+ 5P

1V

2 3P

1V

2+ 3P

2V

2]

=2

1[P

2V

2 P

2V

1 P

1V

2+ P

1V

1 5P

1V

1+ 5 P

1V

2 3P

1V

2+ 3P

2V

2]

=2

1[ 4P

1V

1+ 4P

2V

2+ P

1V

2 P

2V

1]

Q = 2 [(P2V

2 P

1V

1) +

2

1(P

1V

2 P

2V

1]

99. V K [ V (

0mv2

12 2 2+

,

-./

0= 0 + 0 +

,

-./

0

d

Kqq

d = ++

,

-

.

.

/

0

>Z 2

2

0 mv

q.

4

1= 2

0

2

mv4

q

>Z

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100. N0 [ Initial nucleon

at t = 0 J K7N

0N

Add. at a constant rate (7N C) =dt

dN3

@

k

0

dt

= @+

,

-.

/

0

37

N

N0 CN

dN

t =NN0

)CN(n1

377

3

t = ++

,

-

.

.

/

0

37

37

73

CN

CNn

1

0

( e7t = CN

CN

0 37

37

7N C = e7t (7N0 C)

N =7

27

73 teC(7N

0 C) =

7

C+ N

0e7t

7

C.e(7t)

=7C [1 e(7t)] + N

0edt .

CHEMISTRY

101. H2C

2O

4.2H

2O (GMM = 126)

Molarity (M) =100126

100052.2

A

A= 0.2M

M1V

1= M

2V

2

0.2 10 = M2 500

M2

= 4 103.

valence factor of H2C

2O

4.2H

2O = 2.

Normality = [M] V.F. = 2 [4 10 3] = 8 103 N

Amount of oxalic acid (mg/mL) =3

33

10

10126104 A3

+,

-./

0

mL

mg= 0.504.

Note : Options are not match with solution.

102. J K\H J K

]Br

J KHBr + CH

3OH

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103.

104. A J K B

9H = 7.5 kJ/mole ; 9S = 25 J/mole.

9G = 9 H T9S at equilibrium 9 G = 0

0 = 7.5 103 T(25)

T =25

105.7 3A= 300 K

105. [Mg2+] = 102 M

)2)OH(Mg(SPK = [Mg

2+] [OH]2 = 1.0 1012

= [OH]2 =2

12

10

100.13

3A

= 1010

[OH] = 1 105

pOH = 5 ; pH = 9

106. For FCC unit cell

4r = a2

a = ++,

-../

0 A

2

4.1414= 400 pm

Volume of unit cell a3 = (400)3 = 6.4 107 pm3.

107. Formula of cyclic silicate : [SinO

3n]2n ; Cyclic silicates : (SiO

32)

nor (SiO

3)

n2n

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CLASS XII STREAM SB 2 PAGE # 23

108. Target eq., 2B (s) + 3H2(g) K B

2H

6(g)

Given that H2O () K H

2O (g) 9H

10 = 44 kJ .....(i)

2B + 3/2 O2

(g) K B2O

3(s) 9H

20 = 1273 kJ .....(ii)

B2H

6(g) + 3 O

2(g) K B

2O

3(s) + 3 H

2O (g) 9H

30 = 2035 kJ .....(iii)

H2(g) + 1/2 O

2(g) K H

2O () 9H

40 = 286 kJ .....(iv)

Target eq. : Eq.(ii) Eq.(iii) + 3[Eq.(iv)] + 3 [Eq.(i)]

= 1273 [ 2035] + 3[286] + 3[44] = 36 KJ/mole.

109. K3[Fe(CN)

6]

3 + x 6 = 0 x = + 3

26Fe = 3d6 4s2

26Fe3+( d5 with S.L. ( t

2g2,2,1, eg0,0 CFSE = 2.0 9

0+ 2P ~ 2.0 9

0

Number of unpaired electron = 1.

So, B = )2n(n 2 = 3 B.M

110. 9Tf= K

f molality

= 0.925 = 1.85 !"

#$%

&

A

A

92M

10008

M = 173.9 ~ 174Closest answer = 160.

* * * * *

kvpy sol-20101

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