Subject

Chemistry

Class

NEET Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

1.

A hydrogen gas electrode is made by dipping platinum wire in a solution of HCl of pH =10 and by passing hydrogen gas through the platinum wire at 1 atm pressure. The oxidation potential of electrode would be

  • 0.0591 V

  • 0.59 V

  • 0.118 V

  • 1.18 V


B.

0.59 V

For Hydrogen electrode, oxidation half reaction is

stack straight H subscript 2 space with 1 space atm below rightwards arrow stack 2 straight H to the power of plus with left parenthesis at space pH space 10 right parenthesis below space plus 2 straight e to the power of minus
If space pH space equals space 10
straight H to the power of plus space equals space 1 space straight x space 10 to the power of negative pH end exponent space equals space 1 space straight x 10 to the power of negative 10 end exponent
From space nernst space equation comma
straight E subscript cell space equals space straight E subscript cell superscript 0 space minus fraction numerator 0.0591 over denominator 2 end fraction space log space fraction numerator left square bracket straight H to the power of plus right square bracket squared over denominator straight p subscript straight H subscript 2 end subscript end fraction
For space hydrogen space electrode comma space straight E subscript cell superscript 0 space equals space 0
straight E subscript cell space equals negative fraction numerator 0.0591 over denominator 2 end fraction space log space fraction numerator left parenthesis 10 to the power of negative 10 end exponent right parenthesis squared over denominator 1 end fraction
0.0591 space straight x space log space 10 to the power of 10
0.0591 space straight x space 10 space equals space 0.591 space straight V

2596 Views

2.

6.02 x 1020 molecules of urea are present in 100 mL of its solution. The concentration of solution is

  • 0.02 M

  • 0.01 M

  • 0.001 M

  • 0.1 M


B.

0.01 M

Given, number of molecules of urea =6.02 x 1020
therefore ,Number of moles
fraction numerator 6.02 space straight x space 10 to the power of 20 over denominator straight N subscript straight A end fraction
equals space fraction numerator 6.02 space x space 10 to the power of 20 over denominator 6.02 space x space space 10 to the power of 23 end fraction space equals space 1 space x 10 to the power of negative 3 end exponent space m o l
v o l u m e space o f space t h e space s o l u t i o n
space equals space 100 space m L space equals space 100 over 1000 L space equals space 0.1 space L
C o n c e n t r a t i o n space o f space u r e a space s o l u t i o n
left parenthesis i n space m o l space L to the power of negative 1 end exponent right parenthesis space equals space fraction numerator 1 space x space 10 to the power of negative 3 end exponent over denominator 0.1 end fraction space equals space m o l space L to the power of negative 1 end exponent

1596 Views

3.

The value of Planck's constant is 6.63 x 10-34 Js. The speed of light is 3 x1017 nms-1 . Which value is closet to the wavelength in nanometer of a quantum of light with frequency of 6 x 1015 s-1 ?

  • 10

  • 25

  • 50

  • 75


C.

50

Given, Planck's constant,
h= 6.63 x10-34
speed of light, c= 3 x1017 nms-1 
Frequency of quanta
v=6 x1015 s-1
Wavelength, λ =?

We know that,

straight v equals straight c over straight lambda

straight lambda space equals straight c over straight v
equals fraction numerator 3 straight x 10 to the power of 17 over denominator 6 space straight x space 10 to the power of 15 end fraction space equals space 0.5 space straight x space 10 squared space nm space equals space 50 space nm

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4.

A reaction having equal energies of activation for forward and reverse reactions has

  • ΔS =0
  • ΔG =0
  • ΔH = 0
  • ΔH = ΔG=ΔS = 0

C.

ΔH = 0

Energy profile diagram for are reaction is as from the figure it is clear that

(Ea)b = (Ea)f +ΔH
[Here (Ea)b = activation energy of backward reaction and (Ea)f = activation energy of forward reaction].
If (Ea)b = (Ea)b = (Ea)f
then ΔH = 0

2213 Views

5.

Which of the following structure is similar to graphite? 

  • BN

  • B

  • B4

  • B2H6


A.

BN

Boron nitride, (BN)x rersembles with graphite in structure as,

2224 Views

6.

Which of these is least likely to act as a lewis base? 

  • CO

  • F-

  • BF3

  • PF3


C.

BF3

Electron rich species are called lewis base. Among the given, BF3 is an electron deficient species, so have a capacity of electrons accepting instead of donating that's why it is least likely to act as a lewis base. It is a lewis acid. 

2159 Views

7.

Dipole-induced dipole interactions are present in which of the following pair?

  • H2O and alcohol

  • Cl2 and CCl4

  • HCl and He atoms

  • SiF4 and He atoms


C.

HCl and He atoms

Dipole-induced dipole interaction is present in the pair in which the first species is polar and the other is non-polar.
H2O and alcohol both are non-polar so there exist dipole-dipole interactions in between them.
Cl2 and CCl4 both are non-polar so there exists induced dipole -induced dipole interactions in between them. Similarly is true for SiCl4 and He atoms pair.

HCl is a polar molecule, whereas He atoms are non-polar, so in between them dipole-induced dipole interactions exist.
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8.

What is the maximum number of electrons that can be associated with the following set of quantum number?
n=3, l =1 and m=-1.

  • 10

  • 6

  • 4

  • 2


D.

2

The orbital of the electron having =3, l =1 and m= -1 is 3pz (as nlm) and an orbital can have a maximum of two electrons with opposite spins.
therefore, 3pz orbital contains only two electrons or only 2 electrons are associated with n=3, l=1, m=-1.

2636 Views

9.

Based on equation
E=-2.178 x 10-18open parentheses straight Z squared over straight n squared close parentheses certain conclusions are written. Which of them is not correct?

  • The negative sign in the equation simply means that the energy of an electron bound to the nucleus is lower than if would be if the electrons were at the infinite distance from the nucleus.

  • Larger the value of n, the larger is the orbit radius

  • Equation can be used to calculate the change in energy when the electron changes orbit

  • For n=1 the electron has a more negative energy than it does for n=6 which means that the electron is more loosely bound in the smallest allowed orbit.


D.

For n=1 the electron has a more negative energy than it does for n=6 which means that the electron is more loosely bound in the smallest allowed orbit.

If n=1,

E1 = - 2.178 x 10-18 Z2 J

If n=6
space straight E subscript 6 end subscript space equals space fraction numerator negative 2.178 space straight x space 10 to the power of negative 18 end exponent over denominator 36 end fraction

equals space 6.05 space straight x space 10 to the power of negative 20 end exponent space straight Z squared space straight J
From the above calculation, it is obvious that electron has a more negative energy than it does for n=6. it means that electron is more strongly bound in the smallest allowed orbit.

880 Views

10.

How many grams of the concentrated nitric solution should be used to prepare 250 mL o 2.0M HNO3 ? The concentrated acid is 70% HNO3.

  • 45.0 g conc. HNO3

  • 90.0 g conc. HNO3

  • 70.0 g conc. HNO3

  • 54.0 g conc. HNO3


A.

45.0 g conc. HNO3

Given, molarity of solution = 2
Volume of solution = 250 mL = 250/1000 = 1/4 L
Molar mass of 

HNO3 = 1+14+3 x 16 = 63 g mol-1
therefore, Molarity

fraction numerator weight space of space HNO subscript 3 over denominator mass space of space HNO subscript 3 space straight x space volume space of space solution space left parenthesis straight L right parenthesis end fraction
therefore comma
Weight space of space HNO subscript 3 space equals space molarity space straight x space space mol. mass space straight x space volume space left parenthesis straight l right parenthesis

equals 2 space straight x space 63 space straight x space 1 fourth straight g space equals space 31.5 space straight g
It space is space the space weight space of space 100 percent sign space HNO subscript 3
But space the space given space acid space is space 70 percent sign space HNO subscript 3

therefore comma
its space weight space equals space 31.5 space straight x space 100 over 70 space straight g space equals space 45 space straight g

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