Chemistry_Today_April_2017_vk_com_stopthepress

Here, d′ = d + 2.4 × 10 –3 m, t = 3 mm = 3 × 10 –3 m
Substituting these values, we get
−3 −3⎛
1 ⎞
d= d+ ( 24 . × 10 ) − 3× 10 −
⎝
⎜1
K ⎠
⎟
−3 −3⎛
1 ⎞
or ( 24 . × 10 ) = 3× 10 1−
⎝
⎜
⎠
⎟
K
Solving it, we get K = 5
18. (7)
1⎡
1
19. (c) : k =
t ⎣
⎢
a−
x − 1⎤
a⎦
⎥ for second order reaction.
1 1
kt
a− x
= + a
∵ The given graph between (a – x) –1 and time ‘t’ is
linear.
slope = k = tan q = 0.5 L mol –1 min –1
OA = 1 a = 2 L mol–1
a0.5 mol L –1
For second order reaction, rate of reaction is
proportional to the square of concentration.
Rate = k(a) 2 = 0.5 × 0.5 × 0.5 = 0.125 mol L –1 min –1
20. (c)
21. (c) : w = – pDV = – p(V 2 – V 1 ) = –1(20 – 10) atm dm 3
= – 10 atm dm 3 = – 10 × 101.27 J = – 1012.7 J
DU = q + w = 800 – 1012.7 = – 212.7 J – 213 J
22. (a) : d
Z M
Z d a N A
=
×
= × 3
or
×
3
a × N
M
A
−3 −8 3 23 −1
(2 gcm )(5× 10 cm) (6 × 10 mol )
Z =
−1
= 2
75g mol
Since, the number of atoms per unit cell is 2. It
indicates that the metal has body centred cubic
(bcc) lattice.
For bcc lattice, body diagonal of the unit cell,
4 × atomic radius (r) = 3 × edge length (a)
4r = 3 × 5 Å or r = 3
4 × 5 Å = 2.165 Å
23. (a) : The graph reveals that the solubility of gas P
is lowest. Thus, the value of K H for gas P is highest
because higher the value of K H , lower is the
solubility of the gas.
24. (a, b, c) 25. (b, d)
26. (a,b,c,d) : HBr + KOH KBr + H 2 O ;
volume of the resulting solution will be doubled
and the solution will be neutral (pH = 7).
Hence, [K + ] = [Br – ] = 01 . = 0.05 mol L –1
2
[H 3 O + ] = [OH – ] = 1.0 × 10 –7 mol L –1
27. (b): Bridged ion would generate a pair of
enantiomers.
Cl
H
n-pr
H n-pr Cl
Et
H
H Cl
Cl
Et
28. (b,d) : When hard water is passed through zeolite,
Ca 2+ and Mg 2+ react with sodium zeolite and form
calcium and magnesium zeolites.
Na 2 Al 2 Si 2 O 8 . xH 2 O + Ca 2+ CaAl 2 Si 2 O 8 . xH 2 O
+ 2Na +
Na 2 Al 2 Si 2 O 8 . xH 2 O + Mg 2+ MgAl 2 Si 2 O 8 . xH 2 O
+ 2Na +
29. (c) : Cl group present at o- or p- positions to the
electron withdrawing group is activated towards
nucleophilic substitution reaction. Hence, only
Cl present at the o- or p-position to the NO 2
group will be replaced by —NH 2 group.
30. (a, c, d) 31. (a, b, c, d)
32. (5) : O + 2 , CN, NO, N + 2 and CO + have bond order of 2.5.
33. (5) : Mass of glucose = 120 g
No. of moles of glucose = 120
180 = 0.67
Heat produced after eating 0.67 mol of glucose
= 0.67 × 2880 = 1929.6 kJ
Energy available for muscular work
= 1929. 6 ×
25
= 482. 4 kJ
100
Approximate distance that a person will walk
482.
4
= = 4. 824 km≈
5 km
100
34. (3) : [Co 2 (CO) 8 ] has six terminal and two bridged
CO CO CO
CO
CO groups Co Co , the ratio is 6 : 2
CO CO
CO CO
i.e., 3 : 1. Hence, the value of x is 3.
35. (8) : 8B (g) 8A (g) + C (s)
Initial no. of moles : 4V 2V
RT RT
No. of moles at eq. : 4V 2V
− x + x
RT RT
CHEMISTRY TODAY | APRIL ‘17 81