Subject

Chemistry

Class

NEET Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

1.

How many neutrons are there in nucleus 'X', if it gives 7N14 after two successive β - particle emissions ?

  • 7

  • 10

  • 14

  • 09


D.

09

The atomic number of radioactive element increases by one by the emission of 1 β - particle. Hence, after the emission of 2 β -particles, there will be an increase of 2 units in atomic number.

Hence, 5X14 -2β -particle 7N14

Hence, the number of neutrons in parent element (X) = 14 - 5 = 9


2.

The energy equivalent to 1 amu., is :

  • 931.1 MeV

  • 1.492 x 1013 erg

  • 1000 J

  • 107 erg


A.

931.1 MeV

From Einstein equation, E = mc2

 Energy equivalent to 1 amu

(E)= 1 a.m.u. x (3 x 108 m/s)2

= 1.66 x 10-27 Kg x 9 x 1016 m2/s2

= 14.94 x 10-11 joule

(Kg m2/s2 = joule)

14.94 x 10-111.6 x 10-13MeV

= 931.1 MeV


3.

According to Bohr, the angular momentum of an electron, in any orbit, should be :

  • h/2π

  • nh/2π

  • h/mv

  • 2π/nh


B.

nh/2π

According to Bohr's atomic model, electrons can revolve only in those orbits, where its angular momentum is a multiple of h/2π ie equal to nh/2π (where n is an integer)


4.

Which particle can be used to change 13Al27 into 15P30 ?

  • Neutron

  • α - particle

  • Proton

  • Deuteron


B.

α - particle

α - particle can be used to convert 13Al27 into 15P30 .

13Al272He4  15P300n1


5.

The pH is less than 7, of the solution of :

  • FeCl3

  • NaCN

  • NaOH

  • NaCl


A.

FeCl3

FeCl3 is a salt of strong acid and weak base, hence on hydrolysis, it gives acidic solution. Hence, the pH value of solution will be less than 7


6.

pH of a solution of 10 mL 1N sodium acetate and 50 mL 2N acetic acid (Ka = 1.8 x 10-5), is approximately :

  • 4

  • 5

  • 6

  • 7


A.

4

Given, Ka = 1.8 x 10-5

[salt] = 1 x 10 = 10 millimol

[Acid] = 2 x 50 = 100 millimol

From Henderson's equation,

pH = pKa + log [salt][acid]

= - log (1.8 x 10-5) + log 10100

= - log 1.8 + 5 + log 10-1

= - 0.2553 + 5 - 1 = 3.7447

= 4


7.

2.0 molar solution is obtained, when 0.5 mole solute is dissolved in :

  • 250 ml solvent

  • 250 g solvent

  • 250 ml solution

  • 1000 ml solvent


B.

250 g solvent

We know that,

Molality = no. of moles of soluteweight of solvent (in kg)

 2.0 = 0.5weight of solvent (in kg)

 weight of solvent = 0.52.0 = 0.250 Kg = 250 g


8.

The solution obtained by adding water to alcohol, shows :

  • +ve deviation from Raoult's law

  • ideal behaviour

  • -ve deviation from Raoult's law

  • application of Henry's law


A.

+ve deviation from Raoult's law

For the solution of alcohol and water, A ... B attractions, are lesser than B...B and A ... A attractions, hence the vapour pressure of solution is greater than expected. i.e., this solution shows positive deviation from Raoult's law (Vm= +ve, Hm= +ve).


9.

What will be the normality of a solution, containing 4.9 g. H3PO4 dissolved in 500 ml water ?

  • 0.3

  • 1.0

  • 3.0

  • 0.1


A.

0.3

Given that, volume = 500 mL = 0.5 litre

weight of solute (w) = 4.9 g

equivalent weight of

H3PO4molecular weightbasicity

3 × 1 + 31 + 4 × 163 = 983

Hence, normality of solution (N)

wequivalent weight x volume

4.9983×0.5 = 0.3N


10.

The values of Ksp for CuS, Ag2S and HgS are 10-31, 1042, 10-54 respectively. The correct order of their solubility in water is :

  • AgS > HgS > CuS

  • HgS > Cus > Ag2S

  • HgS > Ag2S > Cus

  • Ag2S > CuS > HgS


D.

Ag2S > CuS > HgS

Solubility product (Ksp) of CuS = 10-31

 Solubility of Cus = Ksp = 10-31

= 3.16 x 10-16 mol/lit

Similarly, solubility of Ag2S

3Ksp4

( for Ag2S, 4s3 = Ksp)

10-424

= 6.3 x 10-15 mol/litre

Solubility of HgS = Ksp

10-54

= 10-27 mol/lit

Hence, the correct order of solubility is

Ag2S > Cus > HgS