Subject

Chemistry

Class

NEET Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

1.

In which of the following reactions, standard reaction entropy changes (ΔSo) is positive and standard Gibb's energy change (ΔGo) decreases sharply with increasing temperature?

  • C (graphite) +1/2 O2 (g) → CO (g)

  • CO (g) +1/2 (g) → CO2 (g)

  • Mg(s)  +1/2O2 (g) → MgO (s)

  • Mg(s)  +1/2O2 (g) → MgO (s)


A.

C (graphite) +1/2 O2 (g) → CO (g)

Among the given reactions only in te case of 
C (graphite) +1/2 O2 (g) → CO (g)
entropy increases because randomness (disorder) increases. Thus, standard entropy change (ΔSo) is positive.
Moreover, it is a combustion reaction and all the combustion reactions are generally exothermic, ie. ΔHo=-ve
We know that
ΔGo = ΔHo-TΔSo

ΔGo = -ve-T(+ve)
Thus, as the temperature increases, the value of ΔGo decreases.

1457 Views

2.

pH of a saturated solution of Ba(OH)2 is 12. The value of solubility product Ksp of Ba(OH)2 is

  • 3.3 x 10-7

  • 5.0 x 10-7

  • 4.0 x 10-6

  • 4.0 x 10-6


B.

5.0 x 10-7

Given, pH of Ba(OH)2 = 12
pOH = 14-pH
= 14-12 = 2
We know that,
pOH = -log [OH-]
2 =-log [OH-]
[OH-] = antilog (-2)
[OH-] = 1 x 10-2

Ba(OH)2dissolves in water as 

stack Ba left parenthesis OH right parenthesis subscript 2 space left parenthesis straight s right parenthesis with straight s space mol space straight L to the power of negative 1 end exponent below space rightwards harpoon over leftwards harpoon space stack Ba to the power of 2 plus end exponent with straight s below space plus stack 2 OH to the power of minus with 2 straight s below
left square bracket OH to the power of minus right square bracket space equals space 2 straight s space equals space 1 space straight x space 10 to the power of negative 2 end exponent
left square bracket Ba to the power of 2 plus end exponent right square bracket space equals space fraction numerator left square bracket OH to the power of minus right square bracket over denominator 2 end fraction space equals space fraction numerator 1 space straight x space 10 to the power of negative 2 end exponent over denominator 2 end fraction
straight K subscript sp space equals space left square bracket Ba to the power of 2 plus end exponent right square bracket left square bracket OH to the power of minus right square bracket squared
space equals space open parentheses fraction numerator 1 space straight x space 10 to the power of negative 2 end exponent over denominator 2 end fraction close parentheses left parenthesis 1 straight x 10 to the power of negative 2 end exponent right parenthesis squared
space equals space 0.5 space straight x space 10 to the power of negative 6 end exponent space equals space 5 space straight x space 10 to the power of negative 7 end exponent

3319 Views

3.

The correct set four quantum number for the valence electron of rubidium atom (z=37) is 

  • 5,1,1, +1/2

  • 6,0,0,+1/2

  • 5,0,0 +1/2

  • 5,0,0 +1/2


C.

5,0,0 +1/2

37Rb  = 36[Kr] 5s1
Its valence electron is 5s1
so,
n=5
l=0 (for s orbital)
m = 0 ( As m= -l to +l)

s =+1/2

816 Views

4.

Maximum number of electrons in a subshell with l =3 and n=4 is

  • 14

  • 16

  • 10

  • 10


A.

14

n represents the main energy level and l represents the subshell.
If n=4 and l = 3, the subshell is 4f.
If f subshell, there are 7 orbitals and each orbital can accommodate a maximum number of electrons, so,  maximum number of electrons in 4f subshell = 7 x 2 = 14

798 Views

5.

Standard enthalpy of vaporisation ΔvapHθ for water for water at 100oC is 40.66 kJ mol-1. The internal energy of vaporisation of water at 100o C (in KJ mol-1) is

(Assume water vapour to behave like an ideal gas.)

  • +37.56

  • -43.76

  • +43.76

  • +43.76


A.

+37.56

straight H subscript 2 straight O space rightwards arrow with 100 to the power of straight o space straight C on top space straight H subscript 2 straight O space left parenthesis straight g right parenthesis
increment subscript Vap straight H to the power of straight theta space equals increment subscript vap straight E to the power of straight theta space plus increment straight n subscript straight g RT
For space the space above space reaction comma
increment straight n subscript straight g space equals space straight n subscript straight p minus straight n subscript straight r space equals 1 minus 0 space equals 1
therefore comma
40.66 space KJ space mol to the power of negative 1 end exponent space equals space increment subscript vap straight E to the power of straight theta space plus 1 space straight x space 8.314 space straight x space 10 to the power of negative 3 end exponent space straight x space 373
increment subscript vap straight E to the power of straight theta space equals space 40.66 space KJ space mol to the power of negative 1 end exponent space minus 3.1 space KJ space mol to the power of negative 1 end exponent
space equals plus 37.56 space KJ space mol to the power of negative 1 end exponent
594 Views

6.

50 mL of each gas A and of gas B takes 150 and 200 s respectively for effusing through a pin hole under the similar conditions. If molecular mass of gas B is 36, the molecular mass of gas A will be

  • 96

  • 128

  • 20.2

  • 20.2


C.

20.2

Given,
VA =VB = 50 mL
TA = 150 s
TB = 200 s
MB = 36
MA = ?
From Graham's law of effusion

fraction numerator straight r subscript straight B over denominator straight r subscript straight A space end fraction space equals space square root of M subscript A over M subscript B end root space equals space fraction numerator V subscript B T subscript A over denominator T subscript B. V subscript A end subscript end fraction
rightwards double arrow space square root of M subscript A over 36 end root space equals space fraction numerator V subscript A x 150 over denominator 200 space x space V subscript A end fraction
O r space square root of M subscript A over 36 end root space equals space 15 over 20 space equals space 3 over 4
M subscript A over 36 space equals space 9 over 16
M subscript A space equals space fraction numerator 9 space x space 36 over denominator 16 end fraction space equals space fraction numerator 9 space x space 9 space over denominator 4 end fraction space equals space 81 over 4 space equals space 20.2 space

1395 Views

7.

Buffer solutions have constant acidity and alkalinity because

  • these give unionised acid or base on reaction with added acid or alkali

  • acids and alkalies in these solutions are shielded from attack by other ions.

  • they have a large excess of H+ or OH- ions

  • they have a large excess of H+ or OH- ions


A.

these give unionised acid or base on reaction with added acid or alkali

If a small amount of an acid or alkali is added to a buffer solution, it converts them into unionised acid or base. Thus, remains unaffected or in other words its acidity/alkalinity remains constant. e.g.,

H3O+ + A-  ⇌ H2O +HA

-OH +HA → H2O +A-
If acid is added, it reacts with A- to form undissociated HA. Similarly, if base/alkali is added, O H- combines with HA to give H2O and A- and thus, maintains the acidity/alkalinity of buffer solution.

574 Views

8.

Equimolar solutions of the following substances were prepared separately, which one of these will record the highest pH value?

  • BaCl2

  • AlCl3

  • LiCl

  • LiCl


A.

BaCl2

BaCl2 is a salt of strong acid HCl and strong base Ba(OH)2. So, its aqueous solution is neutral with pH 7. All other salts give acidic solution due to cationic hydrolysis, so their pH is less than 7. Thus, pH value is highest for the solution of BaCl2.

664 Views

9.

Bond order of 1.5 is shown by

  • O2+

  • O2-

  • O22-

  • O22-


B.

O2-

MO configuration of O2+ (8+8-1=15)

straight sigma 1 straight s squared comma space straight sigma asterisk times 1 straight s squared comma space straight sigma 2 straight s squared comma straight sigma asterisk times 2 straight s squared comma straight sigma 2 straight p subscript straight z superscript 2 comma space straight pi 2 straight p subscript straight x superscript 2 almost equal to straight pi 2 straight p subscript straight y superscript 2 space comma space straight pi asterisk times 2 straight p subscript straight x superscript 1 almost equal to straight pi asterisk times 2 straight p subscript straight y superscript 0
Bond space order space space equals space fraction numerator straight N subscript straight b minus straight N subscript straight a over denominator 2 end fraction
left parenthesis where space straight N subscript straight b space equals space number space of space electrons space in space bonding space molecular space orbital
straight N subscript straight a space equals space number space of space electrons space in space anti minus bonding space molecular space orbital right parenthesis

therefore space BO space equals space fraction numerator 10 minus 5 over denominator 2 end fraction space equals space 2.5
Simiarly comma
straight b right parenthesis space straight O subscript 2 superscript minus space left parenthesis 8 plus 8 plus 1 space equals space 17 right parenthesis
so space Bond space order space equals space fraction numerator straight N subscript straight b space minus space straight N subscript straight a over denominator 2 end fraction space equals fraction numerator space 10 minus 7 over denominator 2 end fraction space equals space 1.5

straight c right parenthesis space straight O subscript 2 superscript 2 minus end superscript space left parenthesis 8 plus 8 plus 2 equals 18 right parenthesis
BO space equals space fraction numerator straight N subscript straight b minus straight N subscript straight a over denominator 2 end fraction space equals space fraction numerator 10 minus 8 over denominator 2 end fraction space equals 1

straight d right parenthesis space straight O subscript 2 space left parenthesis 8 plus 8 equals 16 right parenthesis
BO space equals space fraction numerator 10 minus 6 over denominator 2 end fraction space equals space 2
Thus space straight O subscript 2 superscript minus space shows space the space bond space order space 1.5. space

676 Views

10.

The enthalpy of fusion of water is 1.435 Kcal/mol. The molar entropy change for the melting of ice of at 0o C is

  • 10.52 cal/(mol K)

  • 21.04 cal/(mol K)

  • 5.260 cal/ (mol K)

  • 5.260 cal/ (mol K)


C.

5.260 cal/ (mol K)

Molar entropy change for the melting of ice,

ΔS subscript melt end subscript space equals space fraction numerator increment straight H subscript fusion over denominator straight T end fraction
equals space fraction numerator 1.435 space kcal divided by mol over denominator left parenthesis 0 plus 273 right parenthesis space straight K end fraction
equals space 5.26 space straight x space 10 to the power of negative 3 end exponent space kcal divided by mol space straight K
equals space 5.26 space cal divided by mol space straight K

861 Views