Subject

Chemistry

Class

NEET Class 12

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

 Multiple Choice QuestionsMultiple Choice Questions

1.

20 mL 0.1 (N) acetic acid is mixed with 10 ml  0.l(N) solution of NaOH. The pH of the resulting solution is (pKb of acetic acid is 4.74)

  • 3.74

  • 4.74

  • 5.74

  • 6.74


B.

4.74

CH3CHO         +    NaOH

initial 20x0.1       =10x0.1

=2mmol             =1mmol

at time t             =(1-1)

=(2-1)                =0 mmol

=1mmol

 CH3COONa1 mmol + H2O

From Henderson's equation,

pH =pKa + log[CH3COONa][CH3COOH]

=4.74 + log11

=4.74


2.

The equivalent weight of K2Cr2O7 in acidic medium is expressed in terms of its molecular weight (M) as:

  • M/3

  • M/4

  • M/6

  • M/7


C.

M/6

In acidic medium,

K2Cr2O7+6  Cr3+

Equivalent weight of

K2Cr2O7Molecular weightChange in oxidation state of Cr

M2(6-3) = M6


3.

Which of the following is correct?

  • Radius of Ca2+ < Cl- < S2-

  • Radius of Cl- < S2- < Ca2+

  • Radius of Cl- = S2- = Ca2+

  • Radius of S2- < Cl- < Ca2+


A.

Radius of Ca2+ < Cl- < S2-

For isoelectronic species, an increase in atomic number, result in decreased size. Thus, the order of radius will be

Ca2+ < Cl- < S2-

SPECIES ATOMIC NUMBER NUMBER OF ELECTRONS
Ca2+ 20 20-2 = 18
Cl- 17 17+1 = 18
S2- 16 16+2 = 18

4.

11Na24 1s radioactive and it decays to:

  • 9F20 and α-particles

  • 13Al24 and positron

  • 11Na23 and neutron

  • 12Mg24 and β-particles


D.

12Mg24 and β-particles

11Na24  12Mg24 (stable) + -1β0


5.

The state of hybridization of the central atom and the number of lone pairs over the central atom in POCl3
are:

  • sp , 0

  • sp2 , 0

  • sp3 , 0

  • dsp2 , 1


C.

sp3 , 0

σ bonds ⇒ sp3 hybridization without lone pair of electrons.


6.

Which of the following is correct?

  • Evaporation of water causes an increase in disorder of the system

  • Melting of ice causes a decrease in the randomness of the system

  • Condensation of steam causes an increase in disorder of the system

  • There is practically no change in the randomness of the system when water is evaporated


A.

Evaporation of water causes an increase in disorder of the system

Water evaporates into steam. In this physical state its disorder or randomness increases.


7.

In a reversible chemical reaction at equilibrium, if the concentration of any one of the reactants is doubled, then the equilibrium constant will:

  • also be doubled

  • be halved

  • remains the same

  • becomes one-fourth


D.

becomes one-fourth

The equilibrium constant is independent of the concentration of reactants.


8.

Which of the following does not represent the mathematical expression for the Heisenberg uncertainty principle?

  • . h/(4π)

  • . ∆v  h/(4πm)

  • ∆E . ∆t  h/(4π)

  • ∆E . ∆x  h/(4π)


C.

∆E . ∆t  h/(4π)

From the Heisenberg uncertainty principle,

.  h/(4π)      ...(i)

or . m∆v  h/(4π)

or . ∆v  h/(4πm)  ...(ii)

This principle is also applicable for pairs like energy-time (∆E . ∆t) and angular moment-angle (∆w . Θ) along with position-moment (∆x . ∆p).

Thus, ∆E . ∆t  h/(4π)   ...(iii)


9.

A 100 mL 0.1 = (M) solution of ammonium acetate is diluted by adding 100 mL of water. The pH of the resulting solution will be (pKa of acetic acid is nearly equal to pKb of NH4OH).

  • 4.9

  • 5.0

  • 7.0

  • 10.0


C.

7.0

Ammonium acetate is a salt of a weak acid and a weak base. When solutions of such salts diluted resulting pH of the solution is calculated from the following relationship,

pH = 7 + 1/2 pKa - 1/2 pKa

 PKapKb           (given)

 pH of diluted solution = 7

 


10.

Equal weights of CH3 and H2 are mixed in an empty container at 25°C. the fraction of the total pressure exerted by H2 is:

  • 1/9

  • 1/2

  • 8/9

  • 16/17


C.

8/9

The ratio of the number of moles of CH4 and H2 are:

CH4:H2x16:x2 = 1:8

Hence, partial pressure exerted by

H281+8 X Ptotal = 8/9 X ptotal