IIT-JEE 2011 - Career Point

58.5 g of NaCl yields = 143.5 g AgCl
143.5
∴ x g of NaCl yields = x g AgCl
58.5
And from eq. (4),
74.5 g of KCl yields = 143.5 g of AgCl
∴ (0.1180 – x)g of KCl yields
143.5
= (0.1180 – x)g AgCl
74.5
Total mass of AgCl
143.5 143.5
x + (0.1180 – x) = 0.2451
58.5 74.5
which gives, x = 0.0342
Hence, Mass of NaCl = x = 0.0342 g
And Mass of KCl = 0.1180 – 0.0342 = 0.0838g
Step 2. From eq.(1),
117 g of NaCl is obtained from = 62 g Na 2 O
∴ 0.0342 g NaCl is obtained from
62
= × 0.032 = 0.018 g Na2 O
117
From eq. (2),
149 g of KCl is obtained from = 94 g K 2 O
∴ 0.0838 g of KCl is obtained from
94
= × 0.0838 = 0.053 g K2 O
149
0.018
Step 3. % of Na 2 O in feldspar = × 100 = 3.6%
0.5
0.053
% of K 2 O in feldspar = × 100 = 10.6 %
0.5
7. A metallic element crystallizes into a lattice
containing a sequence of layers of ABABAB ..... Any
packing of spheres leaves out voids in the lattice.
What percentage by volume of this lattice is empty
space ?
[IIT-1996]
Sol. A unit cell of hcp structure is a hexagonal cell, which
is shown in fig. Three such cells form one hcp unit.
For hexagonal cell, a = b ≠ c; α = β = 90º and
γ = 120º. It has 8 atoms at the corners and one inside,
hence
Number of atoms per unit cell = 8
8 + 1 = 2
O
a
60º
N b
3
Area of the base = b × ON = b × a sin 60º = a
2
2
( Q b = a)
Volume of the hexagonal cell
3
= Area of the base × height = a 2 . c
2
But c =
2 2
a
3
c
β α
b
a γ
∴ Volume of the hexagonal cell
3
= a 2 2 2
. a = a 3 2
2 3
and radius of the atom,
r = a/2
Hence, fraction of total volume of atomic packing
Volume of 2 atoms
factor =
Volume of the hexagonal cell
4 4 ⎛ a ⎞
3
2×
πr
2 × π⎜
⎟
=
3 3 2
=
⎝ ⎠ π
=
3
3
a 2 a 2 3 2
= 0.74 = 74%
∴ The percentage of void space = 100 – 74 = 26%
8. A basic nitrogen compound gave a foul smelling gas
when treated with CHCl 3 and alc. KOH. A 0.295 g
sample of the substance, dissolved in aq. HCl and
treated with NaNO 2 solution at 0ºC liberated a
colourless, odourless gas whose volume corresponds
to 112 ml at STP. After the evolution of gas was
completed, the aq. solution was distilled to give an
organic liquid which did not contain nitrogen and
which on warming with alkali and I 2 gave a yellow
precipitate. Identify the original sustance. Assume
that it contains one N atom per molecule. [IIT-1993]
Sol. As the compound on heating with CHCl 3 and alc.
KOH gives foul smelling gas, it should be any
primary amine.
RNH 2 + CHCl 3 + 3KOH
3
⎯⎯→
∆ RN C
(Alkyl isocyanide
(foul smelling gas)
+ 3KCl + 3H 2 O
Since the compound on treating with NaNO 2 and HCl
at 0ºC produces a colourless gas, the compound must
be a p-aliphatic amine, because if it was aromatic
diazonium salt might have been produced
XtraEdge for IIT-JEE 13 MAY 2010