NEW CRISTAL ACADEMY, PALAKKAD
First Year Theory - HSE
Model Test No.002 – Chemistry Solutions
Date: 01-September-2013
Topics
Some Basic Concepts in Chemistry
Structure of Atom
1. (d)
2. (a)
3. (c)
4. (d)
5. (d)
6. a = 0.1,
b = 2.24 litre, c = 32,
d = 22.4, e = 2.8,
f = 0.1
7. (i) Law of multiple proportion.
(ii) When two elements combine to form
2 or more compounds, masses of one of
the elements which combines with a fixed
mass of the other bear a simple whole
number ratio.
8.
Homogenous Heterogeneous
Distilled water Muddy water
Iron
Salt solution
9. (i) 2H2 + O2 → 2H2O
4g 32g 36g
Ie to equation, 4g H2 require = 32g O2
∴ 3g H2 require = x g O2
X=
= 24g O2
Hence H2 is the limiting reactant.
(ii) 4g of H2 gives = 36g H2O
∴ 3g of H2 gives = xg H2O
(iii) Oxygen left behind = 29 – 24 = 5g of
O2.
10.
Eleme
nt
% of
ele
me
nts
H
4.07%
C
24.27%
Cl
71.65%
E.F = H2CCl
M.F = (E.F)n
n=
Relative no.
of atoms
Simple
ratio
M.F = (H2CCl)2 = H4C2Cl2.
11. (a) No
(b) It is the no. of moles of solute per litre
of the solution.
(c)
12. (a) → visible
(b) → dumb – bell
(c) →
(d) → Electron.
13. (a). (i) could explain the stability of an
atom
(ii) could explain the atomic spectra of
hydrogen.
(b). (i) Inability to explain the line spectra
of multi electron atom
(ii) Inability to explain Heisenberg
uncertainity principle.
14.
n l m s
Na 3 0 0 -1/2
Al 3 1 1 +1/2
Ga 4 1 1 +1/2
Li 2 0 0 -1/2
15. (a) 29Cu and 24Cr
(b) 29Cu = 1s2, 2s2, 2p6, 3s2, 3p6, 4s1, 3d10
24Cn = 1s2, 2s2, 2p6, 3s2, 3p6, 4s1, 3d5
(c) This is due to the fact that exactly half
filled or full filled orbitals (i.e., d5, d16)
have lower energy and hence have extra
stability.
16. (a) (ii)
(b) Hund’s rule
(c) Pairing of electrons does not take place
until there is a vacant orbital.