cbse 11th chemistry | Online Objective Test

Q1. Find energy of the photon which has wavelength of 0.50 Å. (Hint: h = Planck’s constant = 6.626 ×

10^{- 34}

A. 0.98 ×

10^{- 15}

B. 2.98 ×

10^{- 15}

C. 1.98 ×

10^{- 15}

D. 3.98 ×

10^{- 15}

Answer : Option D
Explaination / Solution:

Energy (E) of a photon having wavelength (λ) is given by the expression, $E = \frac{h c}{λ}$

where , h = Planck’s constant = $6.626 \times 10^{- 34} J s$

c = velocity of light in vacuum = $3 \times 10^{8} m / s$

Wavelength $λ = 0.50 A^{°} = 0.50 \times 10^{- 10} m$

Substituting the values in the given expression of E:

$\frac{6.626 \times 10^{- 34} \times 3 \times 10^{8}}{0.50 \times 10^{- 10}}$ $= 3.98 \times 10^{- 15} J$

Environmental Chemistry Prepare / Learn

Q2. Global warming will not result in:

A. Increasing sea levels

B. Unpredictable climate patterns

C. Increasing the size of the hole in the ozone layer

D. Melting of the ice caps

Answer : Option C
Explaination / Solution:

Global warming will not increase the size of the hole in the ozone layer as temperature does not affect the depletion.

Hydrogen Prepare / Learn

Q3. Consider the reactions (A)

H_{2} O_{2} + 2HI \to I_{2} + 2 H_{2} O (B) HOCl + H_{2} O_{2} \to H_{3} O + Cl - - + O_{2}

A. a reducing agent in (A) and oxidising agent in (B)

B. an oxidising agent in (A) and reducing agent in (B)

C. a reducing agent in both (A) and (B)

D. an oxidising agent in both (A) and (B)

Answer : Option B
Explaination / Solution:

H_{2} O_{2}

is an oxidizing agent in 1st reaction and Reducing agent in 2nd reaction.

Thermodynamics Prepare / Learn

Q4. for the reaction at 298 K,

2A + B \to C

= 400 kJ

mo l^{- 1}

and

Δ S

= 0.2 kJ

K^{- 1} mo l^{- 1}

At what temperature will the reaction become spontaneous considering

Δ H

and

Δ S

to be constant over the temperature range.

A. 3500 K

B. 2000 K

C. 1500 K

D. 2500 K

Answer : Option B
Explaination / Solution:

ΔG= ΔH-TΔS At equilibrium ΔG=0 then T= ΔH/ΔS =2000K

Chemical Bonding and Molecular Structure Prepare / Learn

Q5. The

s p^{3} d^{2}

hybridization of central atom of a molecule would lead to

A. Tetrahedral geometry

B. Trigonal bipyramidal geometry

C. square planar geometry

D. Octahedral geometry

Answer : Option D
Explaination / Solution:

In sp3d2 Hybridization one s, three p and two d-orbitals get hybridized to form six sp3d2 hybrid orbitals which adopt octahedral arrangement as given in image below

The s-Block Elements Prepare / Learn

Q6. Radium salts are used in

A. in the treatment of infectious diseases

B. in the treatment of metal sheets

C. in the treatment of tuberculosis

D. in the treatment of cancer

Answer : Option D
Explaination / Solution:

Ra salts are used in the treatment of Cancer.

Organic Chemistry Some Basic Principles and Techniques Prepare / Learn

Q7. Geometrical isomers and Optical isomers are:

A. position isomerism.

B. chain isomers

C. stereoisomer

D. functional isomers

Answer : Option C
Explaination / Solution:

They are stereoisomers as they differ in the orientation of atoms in space.

States of Matter Prepare / Learn

Q8.

Pay load is defined as the difference between the mass of displaced air and the mass of the balloon. Calculate the pay load when a balloon of radius 10 m, mass 100 kg is filled with helium at 1.66 bar at

27^{\circ} C

. (Density of air = 1.2 kg m-3 and R = 0.083 bar

d m^{3} K^{- 1} mo l^{- 1}

A. 3345.2 kg

B. 3811.1 kg

C. 3568.2 kg

D. 3905.3 kg

Answer : Option B
Explaination / Solution:

Volume of balloon,V=4/3(3.14)

r^{3}

=4187

m^{3}

Mass of displaced air,m=V × δ =4187 x 1.2=5024.4 kg Temperature,T=300 K Moles of gas present,n=PV/RT=279.1 × 103 moles Mass of helium=279.1 ×103 x 4=116.4 kg Mass of filled balloon,m'=100+1116.4=1216.4 kg Payload=m-m'=3811.1 kg

Redox Reactions Prepare / Learn

Q9. Using the standard electrode potential, find out the pair between which redox reactions is not feasible.

E 0 v a l u e s : F e 3 + / F e 2 + = + 0.77; I 2 / l^{-} = + 0.54

C u^{2 +} / Cu = + 0. 34; A g^{+} / Ag = + 0. 80 V

A. Fe3+and Cu

F e^{3 +}

and

l^{-}

C. Ag+ and Cu

D. Ag and

F e^{3 +}

Answer : Option D
Explaination / Solution:

As

E_{c e l l}

for this couple is negative so this couple is not possible.

Some Basic Concepts of Chemistry Prepare / Learn

Q10. What is the mass of 1 mole of O2 ?

A. 8.0 g

B. 32.0 g

C. 6.02 g

D. 16.0 g

Answer : Option B
Explaination / Solution:

Calculations:

Mass of 1mole of O2

= the mass of 6.022 x 1023 molecules of O2 ' ( as per Avogadro's law )

= molar mass of O2 in grams ( ie. gram molar mss )

= 32 .0 g

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