Subject

Chemistry

Class

NEET Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

11.

An atomic nucleus having low n/p ratio tries to find stability by

  • the emission of an α-particle

  • the emission of a positron

  • capturing an orbital electron (K-electron capture)

  • emission of β -particle


B.

the emission of a positron

Emission of a positron results in the conversion of a proton into neutron, along with the release of neutrino. Thus, the number of protons decreases and of neutron increases, i.e., n/p ratio increases and the nuclei becomes stable.

H11Proton  n01Neutron + e0+1Psitron + v


12.

Given that,

    C + O2 → CO2H = -x kJ

2CO + O2 → 2CO2 ;H = -y kJ

The heat of formation of carbon monoxide will be

  • y - 2x2

  • y + 2x

  • 2x - y

  • 2x - y2


A.

y - 2x2

Given,

    C + O2 → CO2H° = -x kJ        ...(i)

2CO + O2 → 2CO2H° = -y kJ      ...(ii)

Required equation as per question is C + 12O2 → CO, H1°

On reversing equation (ii), we get

2CO2 → 2CO + O2H2° = +y kJ      ...(iii)

Dividing equation (iii) by 2 gives the following reaction

CO2 → CO + 12O2 ; H3° = + y2kJ    ...(iv)

On adding equation (i) and (iv), we get 

C + 12O2 → CO ; H1° = H° + H3°          = -x + y2          = y - 2x2 kJ


13.

In case of heteronuclear diatomics of the type AB, where A is more electronegative than B, bonding molecular orbital resembles the character of A more than that of B. The statement 

  • is false

  • is true

  • cannot be evaluated since data is not sufficient

  • is true only for certain systems


B.

is true

The given statement is true as electrons are shifted more towards the more electronegative atom.


14.

(32Ge7634Se76) and (14Si3016S32) are examples of

  • isotopes and isobars

  • isobars and isotones

  • isotones and isotopes

  • isobars and isotopes


B.

isobars and isotones

32Ge76 and 34Se76 have the same mass number but different atomic number, therefore, they are isobars.

Number of neutrons, in

 14Si30 = 30 - 14 = 16

16Si32 = 32 - 16 = 16

Because of the presence of same number pf neutrons, 14Si30 and 16S32, they are known as isotones.


15.

The pH of 10-4M KOH solution will be

  • 4

  • 11

  • 10.5

  • 10


D.

10

10-4 M KOH = 10-4 M [OH-]

We know that,

[H+][OH-] = 1 × 10-14

[H+] × 10-4 = 1 × 10-14

[H+] = 1 × 10-141 × 10-4 = 1 × 10-10 M

 pH = - log [H+] = -log (1 × 10-10 M) = 10


16.

The system that contains the maximum number of atoms is

  • 4.25 g of NH3

  • 8 g of O2

  • 2 g of H2

  • 4 g of He


C.

2 g of H2

Number of atoms = massmolar mass × NA × Number of atoms in 1 mole

Number of atoms in 8 gm

O2832  × NA × 2 = NA2

Number of atoms in 2 gm

H222 × NA × 2 = 2NA

Number of atoms in 4 gm

He = 44 × NA × 1 = NA

Thus, 2 gm H2 contains the maximum number of atoms among the given.


17.

The values of H and S of a certain reaction are -400 kJ mol-1 and -20 kJ mol-1K-1 respectively. The temperature below which the reaction is spontaneous, is

  • 100 K

  • 20°C

  • 20 K

  • 120°C


C.

20 K

Given,

H = -400 kJ mol-1 S = -20 kJ mol-1 K-1

Gibbs Helmholtez equation is

G = H - TS

For a reaction to be spontaneous, G must be less than 0, i.e. it should be negative.

 0  H - TS      TS  H      T  HS  -400-20  20 


18.

The value of H for cooling 2 mole of an ideal monoatomic gas from 225°C to 125°C at constant pressure will be [given Cp52R].

  • 250 R

  • -500 R

  • 500 R

  • -250 R


B.

-500 R

The value of H is

H = -nCpT

      = -2 × 52R ( 225 - 125)

      = -5R (100)

      = -500 R

Therefore, H of -500 R is required for cooling 2 mole of an ideal monoatomic gas from 225°C to 125°C at constant pressure.


19.

98Cf246 was formed along with a neutron when an unknown radioactive substance was bomabrded with 6C12. The unknown substance is

  • 91Pa234

  • 90Th234

  • 92U235

  • 92U238


C.

92U235

Supoose, the unknown radioactive substance is xAy.

xAy6C12 → 98Cf2460n1

On comparing atomic numbers, we get

x + 6 = 98 + 0

x = 98 - 6 = 92

On comparing mass numbers, we get

y + 12 = 246 + 1

y = 247 - 12 = 235

Since, uranium has the atomic number 92, thus, the given unknown substance was 92U235.


20.

The enthalpy of vaporisation of a certain liquid at its boiling point of 35°C is 24.64 kJ mol-1. The value of change in entropy for the process is

  • 704 JK-1mol-1

  • 80 JK-1mol-1

  • 24.64 JK-1mol-1

  • 7.04 JK-1mol-1


B.

80 JK-1mol-1

Entropy of vapourisation (S) = enthalpy of vaporisation (H)boiling point (inK)

Given, H = 24.64 kJ mol-1 and boiling point = 35 + 273 = 308 K

 S = 24.64 × 103 J mol-1308 K         = 80 JK -1mol-1