﻿ An atomic nucleus having low n/p ratio tries to find stability by from Chemistry NEET Year 2014 Free Solved Previous Year Papers

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# 11.An atomic nucleus having low n/p ratio tries to find stability bythe 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.

12.

Given that,

C + O2 → CO2$∆$H = -x kJ

2CO + O2 → 2CO2 ;$∆$H = -y kJ

The heat of formation of carbon monoxide will be

• y + 2x

• 2x - y

A.

Given,

C + O2 → CO2$∆\mathrm{H}°$ = -x kJ        ...(i)

2CO + O2 → 2CO2$∆\mathrm{H}°$ = -y kJ      ...(ii)

Required equation as per question is C + $\frac{1}{2}$O2 → CO, $∆{\mathrm{H}}_{1}^{°}$

On reversing equation (ii), we get

2CO2 → 2CO + O2$∆{\mathrm{H}}_{2}^{°}$ = +y kJ      ...(iii)

Dividing equation (iii) by 2 gives the following reaction

CO2 → CO + $\frac{1}{2}{\mathrm{O}}_{2}$ ; $∆{H}_{3}^{°}$ = + $\frac{\mathrm{y}}{2}\mathrm{kJ}$    ...(iv)

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

C + $\frac{1}{2}{\mathrm{O}}_{2}$ → CO ;

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 × 10-10 M

$\therefore$ 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 =  × NA × Number of atoms in 1 mole

$\therefore$Number of atoms in 8 gm

O2

Number of atoms in 2 gm

H2

Number of atoms in 4 gm

He =

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

17.

The values of and $∆\mathrm{S}$ 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

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

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 Cp$\frac{5}{2}\mathrm{R}$].

• 250 R

• -500 R

• 500 R

• -250 R

B.

-500 R

The value of $∆$H is

$∆$H = -nCp$∆\mathrm{T}$

= -2 × $\frac{5}{2}$R ( 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) =

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

$\therefore$