CHETTINAD VIDYASHRAM

R.A. PURAM, CHENNAI-28

CYCLE TEST - I [2021-22]

DATE:26-07-2021 CLASS : XI
TIME: 1 HOUR SUBJECT : CHEMISTRY
MARKS : 30

General Instructions
i. Questions number 1 to 6 carry 1 mark each
ii. Questions number 7 to 11 carry 2 marks each
iii. Questions number 12 to14 carry 3 marks each
iv. Questions number 15 carry 5 marks
v. Use log tables, if necessary, use of calculator is not allowed

1. A sub-shell with n = 6, l = 2 can accommodate a maximum of

(a) 12 electrons
(b) 36 electrons
(c) 10 electrons
(d) 72 electrons
2. Which of the following terms are unitless?
(a) Molality
(b) Molarity
(c) Mole fraction
(d) Mass percent
3. The percentage of Carbon in Ca is
(a) 15%
(b) 1.8%
(c) 14.8%
(d) 15.2%
4. Which of the following statements does not form a part of Bohr’s model of
hydrogen atom?
(a) Energy of the electrons in the orbit is quantised
(b) The electron in the orbit nearest to the nucleus has the lowest energy
(c) Electrons revolve in different orbits around the nucleus
(d) The position and velocity of the electrons in the orbit cannot be
determined simultaneously
5. Which of the following sets of quantum numbers represents the highest
energy of an atom?
(a) n = 3, l = 0, m = 0, s = +
(b) n = 3, l = 1, m = 1, s = +
(c) n = 3, l = 2, m = 1, s = +
(d) n = 4, l = 0, m = 0, s = +
6. 1 Mole of C contains
(a) 6.02× atoms of H
(b) 4 gram-atoms of hydrogen
(c) 1.81× molecules of C
(d) 3g of Carbon

7. The density of 3 molal solution of NaOH is 1.110 g/mL. Calculate the

molarity of the solution.

8. (i) Why molality is preferred over molarity in expressing the

concentration of a solution?
(ii) State Law of Multiple Proportion.

9. Electrons are emitted with zero velocity from a metal surface when it is
exposed to radiation of wavelength 7200Å. Calculate threshold frequency
(v0) and work function(W0) of the metal.

10. 3.0g of hydrogen react with 29.0 g 0f oxygen to yield water.

(i) Which is the limiting reagent?
(ii) Calculate the maximum amount of water that can be formed.
(iii) Calculate the amount of one of the reactants which remain unreacted.

11.(i) We don’t see a car moving as a wave on the road. Why?

(ii) Draw the shapes of (i) (ii)

12.(i)The molecular mass of an organic compound is 78 and its percentage

composition is 92.4% Carbon and 7.6% Hydrogen. Determine the molecular
formula of the compound.
(ii)What is the relation between kinetic energy and frequency of
photoelectrons?
13.(i)Calcium carbonate reacts with aqueous HCl to give and
according to the reaction,
(s) + 2HCl(aq)  + O(l)
What mass of is required to react completely with 50 mL of
 0.75 M HCl? (Atomic mass of Ca = 40u, C= 12u, O=16u)

(ii)Express the following numbers to three significant figures:

(a) 6.023 x (b) 72000

14.Write the electronic configuration of (i) Mn4+, (ii) Cr2+ and(iii) Zn2+
Mention the number of unpaired electrons in each case.
(Z of Mn = 25, Cr =24 and Zn =30)
15.. (i)What transition in the hydrogen spectrum would have the same
wavelength as the Balmer transition n = 4 to n = 2 of spectrum?

(ii) What are degenerate orbitals? Which of the following orbitals are
degenerate?
3 , ,

(iii) Give the mathematical expression of uncertainty principle.

*******************************
 CHETTINAD VIDYASHRAM
R.A. PURAM,
CHENNAI-28

CYCLE TEST - I [2021-22]

ANSWER KEY

CLASS: XI MARKs: 30
SUB: CHEMISTRY

1. (c) 10 electrons (1)

2. (c) Mole fraction (1)
3. (c) 14.8% (1)
4. (d) The position and velocity of the electrons in the orbit cannot be determined
simultaneously (1)
5. (c) n = 3, l = 2, m = 1, s = + ½ (1)
6. (b) 4 gram-atoms of hydrogen (1)
7. Molality of NaOH solution = 3 molal
3moles of NaOH is present in 1000g of the solvent
120g of NaOH is present in 1000g of the solvent (1/2)
Mass of Solution = Mass of Solvent + Mass of Solute

Mass of the solution = 1000 + 120 g = 1120g (1/2)

Density of solution = 1.110g/mL

Volume = Mass/density

= 1120g/1.110
= 1009.0 mL = 1.009 L (1/2)
Molarity = (number of moles of solute)/(volume of the sol: in L)
= 3moles/(1.009 L) = 2.97 mol/L (1/2)

8. Molarity is number of moles per unit volume of the solution and molality is number of
moles per unit mass of solvent. The volume changes with temperature while the mass
of solvent is independent of temperature. So, molality remains constant but molarity
changes with temperature. Hence, molality is preferred over molarity. (1)
(ii) Law of multiple proportions, states that when two elements combine with each other
to form more than one compound, the masses of one element that combine with a fixed
mass of the other element are in a ratio of small whole numbers. (1)

9. Since K.E is zero, the energy of the radiation is equal to the work function
 ν ,ν

The threshold frequency = (1/2)

=
= 4.166 x (1/2)
Work function (1/2)
= 6.626× × 4.166 x
= 2.7603 x J (1/2)
10. 2H₂ + O₂  2H₂ O
(1/2)
4g 32g 36g (1/2)
(i) To determine the limiting reagent.

4 g of H₂ reacts with 32 g of O₂
1 g of H₂ reacts with 32/4 g of O₂
3 g of H₂ reacts with 32/4 x 3 g = 24 g of O₂

But only 29 g of O₂ is present.

So, the limiting reagent is hydrogen. (1/2)

(ii) Now, 4 g of H₂ forms 36 g of H₂ O

1 g of H₂ forms 36/4 g of H₂ O.
3 g of H₂ forms 36/4 x 3 g = 27 g of H₂ O
(1/2)
Maximum amount of water that can be formed is 27 g. (1/2)
(iii) For, amount of oxygen left unreacted,
Only 24 g of oxygen will react.
Amount of oxygen unreacted = 29 - 24 = 5 g (1/2)

11. (i) According to de Broglie's relation, λ = h/mv i.e. λ α 1/m (1)

Mass of the car is very large and its wavelength (λ) or wave character is
negligible. Therefore, we do not see a car moving like a wave

(ii) (1/2, 1/2)

12. (i) (1/2)

Element Percentage Atomic mass Moles of atoms Simple ratio Whole no:
comp: ratio
Carbon 92.4% 12u = 7.7 1

Hydrogen 7.6% 1u = 7.6 =1 1

Empirical formula is CH (1/2)

Empirical formula mass = 12 + 1 = 13

n = Molecular mass/ Empirical formula mass
= 78/13 = 6 (1/2)

Molecular formula = n x Empirical formula

= 6 x CH
Molecular formula is (1/2)

(ii) The kinetic energy of photoelectrons increases with frequency of

radiation (1)

13. (s) + 2HCl  + O

100g (2x 36.5)g

Molar mass of HCl = 36.5g

Molarity = (1/2)
=

= 0.75mol x 36.5g x 50 x L

= 1.368 g (1/2)

2 mol of HCl (2 × 36.5 = 73 g) react with 1 mol of CaCO3 (100 g). (1/2)

Amount of CaCO3 that will react with 1.368g =

= 1.874g (1/2)

ii) (a) 6.02 x

(b) 7.20 x (1/2, 1/2)

14. i) Mn has atomic number 25 and its electronic configuration is
2 2 6 2 6 5 2
1s 2s 2p 3s 3p 3d 4s
2 2 6 2 6 3
Electronic configuration of Mn4+ = 1s 2s 2p 3s 3p 3d (1/2)

= 3 unpaired electron (1/2)

ii) Cr has the atomic number 24 and its electronic configuration
2 2 6 2 6 5 1
1s 2s 2p 3s 3p 3d 4s
2 2 6 2 6 4
Electronic configuration of Cr2+ = 1s 2s 2p 3s 3p 3d (1/2)

= 4 unpaired electron (1/2)

iii) Zn has the atomic number 30 and its electronic configuration is

2 2 6 2 6 10 2
1s 2s 2p 3s 3p 3d 4s
2 2 6 2 6 10 2
Electronic configuration of Zn2+ = 1s 2s 2p 3s 3p 3d 4s (1/2)

= no unpaired electrons (1/2)

15. (i) The wave number for Balmer transition of n =4 to n=2 of He⁺ spectrum would be:
= 4 , Z =2

(1/2)

(1/2)

Acc. to question, the transition for Hydrogen will have the same wavelength: -

(1/2)

This equation can only be true when and (Lyman series) (1/2)

The transition of electron from n = 2 to n = 1 in Hydrogen would have the same

wavelength as transition of electron from n = 4 to n = 2 in
(ii) Degenerate orbitals are orbitals that have the same energy (1)
3 , are degenerate (1/2)
, , are degenerate (1/2)
(iii) or (1/2)

where

- uncertainty in position

- uncertainty in momentum (1/2)

- uncertainty in velocity
h – Planck’s constant

******************