SOLUTIONS
PART B - CHEMISTRY
31.
a)
b)
b)
0.25
50%
Let CaCO3 be x g and SiO2 be (4 – x) g
100 gm of CaCO3 produces 44 gm of CO2,
No. of moles
32.
35.
PCM
5.6
= 0.25
22.4
=
so 0.88 g CO2 is produced by
Ÿ
100
=
x
Ÿ
x =
3.011 u 1023
No. of moles
=
mass
molar mass
=
9
18
=
1
44
0.88
100 u 0.88
44
= 2 gm of CaCO3
so, 50% SiO2 is present.
2
1
No. of molecules =
u 6.022 u 1023
2
= 3.011 u 1023
36.
c)
1.26 u 1024
1 mol C6H12O6 has 6NA atoms of C.
0.35 mol has Ÿ 6 u 0.35 NA = 2.1 N A
atoms of C.
33.
a)
2.1 u 6.022 u 1023
0.01
H2 O
AgNO3 + NaCl 
o AgCl + NaNO3
moles
1.7
170
0.585
58.5
?
0.01
0.01
?
37.
c)
0.85
85
= 1.26 u 1024
26,18
Let atomic mass of P and Q be a and b
respectively.
Molar mass of P2Q3 = 2a + 3b
0.01
Molar mass of P2Q3 = a + 2b
(molar mass u moles = mass)
So (2a + 3b) 0.15 = 15.9
Mass of AgCl formed (by law of mass
conservation)
1.7 + 0.585 – 0.85 = 1.435 gram
1.435
moles of AgCl =
143.5
34.
d)
= 0.01
38.
8.8 g
c)
(principle of atomic conservation)
+ 2HCl(aq) o CaCl2(aq) +
H2O(l) + CO2(g)
for HCl
Ÿ
20
= 0.2 mole
100
a = 26,
b = 18
C7H8
x :
y
=
2
3.08 0.72
:
44
18
= 0.07 : 0.04
§ moles of reactant taken ·
¨
¸ comes out
© stoichiometric coefficient ¹
least for limiting reagent
for CaCO3 Ÿ
= 9.3
y
y·
§
C x Hy ¨ x ¸ O2 o x CO2 H2O
2
4¹
©
1st balance equation using POAC
CaCO3(s)
(a + 2b) 0.15
= 7 : 4
So, x : y
39.
2.0
= 0.548
36.5
b)
2
moles
so CaCO3 is limiting reagent.
1 mole CaCO3 produces 1 mole CO2, so
= 7 : 8
=
molar mass
5.6
=
22.4
=
mass
no.of moles
=
16
1
4
0.2 mole produce 0.2 mole CO2.
mass = molar mass u moles
= 44 u 0.2
= 8.8 gm CO2
1
4
= 64
SOX = 32 + x(16) = 64
... 5 ...
PCM
SOLUTIONS
Ÿ
x = 2
44.
40.
b)
c)
3.125 u 10
Mg3(PO4)2
11.2 kg
–2
Mass of pure Fe2O3 =
8 mole O atoms are in 1 mole
1 mole O atoms in
= 16kg
Fe2O3 o 2Fe
1 mol produces 2 mol
1
mole
8
160g produces 112g
1
1
1 1
mole O atoms in u =
moles
32
4
8 4
= 3.125 u 10–2
1.2
41.
c)
35
2
16kg produces
45.
× NA
d)
=
=
No. of molecules in 1 drop =
2.4
70
% yield =
2.4 1
u
70 35
2.4
35
2
c)
u
NA
2
46.
d)
35
2
Ÿ
C4H10
% wt At. wt. relative no. of
of atoms
6.9y
y6.9 = 1
82.8
12
82.8y
y12 = 6.9
H
17.2
1
17.2y
y1=17.2
47.
17.2y
y6.9=2.5Ÿ
Ÿ5
43.
b)
Ÿ C4H10
10
Na2CO3 . xH2O
Na2CO3 + xH2O
1 mol
1mol
2.574
106 +18x
48.
c)
0.954
106
= 9 u 10–3
= 10
H – F > H – O > H – N > H – Cl
solubility in water (BaSO4 > Na2SO4)
BaSO4 has more covalent character.
49.
b)
Na2O2 > KO2 > O2
Bond order
Bond order decreases.
72 litre
On balancing the equation
4NH3 + 5O2 o 4NO + 6H2O
=
Decreasing electronegativity.
C2H5 Ÿ empirical formula mass = 29
n = 2
n(C2H5)
b)
1
u 123 = 41g
3
24.6
u 100 = 60%
41
On solving x
simple ratio
atoms/moles
C
o C6H5NO2 + H2O
26 2 1
= = mol
78 6 3
1.2 u NA
=
42.
60%
C6H6 + HNO3
= 2.4 gm
No. of moles in 1 drop
112
u 16 = 11.2kg
160
actual yield
% yield = theoretical yield u 100
mass = vol u density
= 2 ml u 1.2 g ml–1
No. of moles in 35 drops
80
u 20
100
50.
Volume ratio = molar ratio
90
5
=
= x = 72
x
4
... 6 ...
c)
NO2 and O3
SOLUTIONS
57.
51.
a)
PCM
a)
X is the limiting reagent
3x + 4y o X3Y4
6
3
BF4
58.
c)
ª6º
« 4 » o minimumso y inLR
¬ ¼
both a and b
1 amu = 1.66x
x10
x –24g
59.
d)
96500 Coloumbs
1.6 x10
x –19coloumb x 6.022 x1023
$ 96500 coloumb/mol.
52.
d)
four 2C – 2e bonds and two 3C – 2e
60.
bonds
d)
both a and b are incorrect
No of atoms on each side remains
conserved (always)
Limiting reagent is decided by both
stoichiometric coefficients of balanced
chemical reaction and moles taken.
53.
d)
CN– and NO+
can be solved by both lewis concept and
MOT
and
MOT, B.O =
=
54.
a)
(B.O = 3)
No of el in BMO – No of el in ABMO
2
10 – 4
=3
2
40
16
u 100 = 40%
40
55.
c)
112 g
88g CO2 contain 64 g oxygen
28g CO contain 16 g oxygen
for 64 g of O, 112 g of CO in needed.
56.
d)
168
C 5 H10 15
O o 5CO2 5H 2O
2 2
15
15
mole Ÿ
u 22.4 = 168 litre
2
2
... 7 ...