Chemistry_Today_April_2017_vk_com_stopthepress

CHEMISTRY MUSING
SOLUTION SET 44
240×
1000
1. (d) : Bond energy per molecule of I 2 =
6.
022×
10 23 J
= 3.985 × 10 –19 J
−34 8
Energy absorbed = hc 6. 626× 10 × 3×
10
=
λ −10
4500×
10
= 4.417 × 10 –19 J
∴ K.E. of one I 2 molecule = (4.417 × 10 –19 – 3.985 × 10 –19 )J
= 4.32 × 10 –20 J
20
432 . × 10
−
K.E. of one I atom = = 2.16 × 10 –20 J
2
2. (d) : Nucleophilic addition reaction to carbonyl
compound takes place followed by intramolecular
nucleophilic substitution reaction.
O
Cl CH3MgBr, dry ether, 0°C
CH 3 Nucleophilic addition reaction
MgBrCl +
O
CH 3
CH 3
aq. acid
Cl
intra molecular
nucleophilic
substitution
reaction
OMgBr
CH
CH 3
3
3. (b) : The colourless inorganic salt (A) is ammonium
nitrate.
NH NO
NO + 2H O
4 3 2 2
( A ) ( B ) ( C)
Product (B) N 2 O is a neutral gas, product (C) H 2 O
is liquid and neutral to litmus.
10N2O + P4 PO
4 10
+ 10N2
O
4. (a) : NH
O
Br 2/KOH
(Hoffmann bromamide
reaction)
(Dehydrating agent)
NaOH
H
HN 2
OOH
ONH
CH 2
CH 2
COOH
-Alanine
(II)
5. (d) : Pb 2+ + 2HCl PbCl 2 ↓ HS 2
PbS ↓ + 2HCl
White ppt. Black ppt.
(dissolves on boiling)
r1
6. (c) : Either octahedral voids = 0. 414
or
r2
r1
tetrahedral voids = 0. 225
are occupied by
r2
[where r 1 is radius of the interstitial site (void)
and r 2 is radius of atoms arranged in fcc]
interstitial sites in fcc.
Since in fcc, atoms along face diagonal are touching,
thus, 4r 2 = 2 a
Required diameter of interstitial sites = 2r 1
= 2 × 0.414 r 2 = 2 × 0 . 414 × 2 a
4
2× 0.
414× 2 × 400
=
4
= 117.1 pm
7. (b) : SnO 2 + 2NaOH Na 2 SnO 3 + H 2 O
2–
SnO 2 + SnO 3
2–
[SnO 2 ] ⋮ SnO 3
As they form negatively charged particles, they are
easily coagulated by AlCl 3 in which Al 3+ cation
carries maximum positive charge.
8. (c) : 50 mL of gold for protection requires
= 0.1 g = 100 mg of starch
∴ 10 mL of gold will require = 20 mg of starch
∴ Thus, gold number of starch = 20
9. (5) : We know, p o
− p n
o =
2
p n1+
n2
Given that: p o = 640 mm Hg, p = 600 mm Hg
Let M be the molecular weight of the solute.
Molar mass of benzene (C 6 H 6 ) = 6 × 12 + 6
= 78 g mol –1
n
.
2 = 175 ;
M
n 1 =
78
640 − 600 2. 175 / M
∴
=
; M = 65.25
640 2.
175
+ 05 .
M
60 + x × 1.05 = 65.25
∴ x = 5
10. (7) : Acidified K 2 Cr 2 O 7 , CuSO 4 , H 2 O 2 , Cl 2 , O 3 ,
FeCl 3 and HNO 3 oxidise iodide to iodine. Alkaline
KMnO 4 oxidises aqueous iodide to IO – 3 ion. Na 2 S 2 O 3
is a strong reducing agent which on reaction with I 2
produces I – ion.
2Na 2 S 2 O 3 + I 2 2NaI + Na 2 S 4 O 6
MPP CLaSS XI ANSWER KEY
1. (c) 2. (b) 3. (a) 4. (d) 5. (d)
6. (c) 7. (c) 8. (d) 9. (a) 10. (c)
11. (b) 12. (c) 13. (c) 14. (c) 15. (c)
16. (d) 17. (a) 18. (d) 19. (d) 20. (a,b,c)
21. (c,d) 22. (a,b,c,d) 23. (b,d) 24. (3) 25. (4)
26. (2) 27. (a) 28. (a) 29. (a) 30. (a)
58 CHEMISTRY TODAY | APRIL ‘17