Subject

Chemistry

Class

NEET Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

21.

In a face-centered cubic lattice, a unit cell is shared equally by how many unit cells?

  • 8

  • 4

  • 2

  • 6


D.

6

In a face centered cubic lattice a unit cell is shared equally by six unit cells.


22.

At 25°C, the dissociation constant of a base, BOH, is 1.0 x 10-12. The concentration of
hydroxyl ions in 0.01 M aqueous solution of the base would be

  • 2.0 x 10-6 mo L-1

  • 1.0 x 10-5 mo L-1

  • 1.0 x 10-6 mo L-1

  • 1.0 x 10-7 mo L-1


D.

1.0 x 10-7 mo L-1

Base BOH is dissociated as follows:

BOH B+ + OH-

So, the dissociation constant of BOH base

Kb[B+] [OH-][BOH]

At equilibrium [B+] = [OH-]

 Kb = [OH]-2[BOH]

Given that, 

Kb = 1.0 x 10-12 and [BOH] = 0.01M

Thus, 1.0 x 10-12[OH-]20.01

[OH-]2 = 1 x 10-14

[OH-] = 1.0 x 10-7 mol L-1


23.

The rate of reaction between two reactants A and B decreases by a factor of 4, if the concentration of reactant B is doubled. The order of this reaction with respect to reactant B is

  • -1

  • -2

  • 1

  • 2


B.

-2

A + B Product

Rate [A] [B]-2            ...(i)

The rate decreases by factor 4, if the concentration of reactant 'B' is doubled.

So, rate [A] [2B]-2

             [A] [B]-24       ...(ii)

Hence, order of reaction wrt reactant B is -2.


24.

The best method for the separation of naphthalene and benzoic acid from their mixture is

  • chromatography

  • crystallisation

  • distillation

  • sublimation


D.

sublimation

The best method for the separation of naphthalene and benzoic acid from their mixture is sublimation because it is applicable for those organic compounds which pass directly from solid to vapour state on heating and vice-versa on cooling. In these compounds naphthalene is volatile and benzoic acid is non-volatile due to the formation of dimer via hydrogen bonding (intermolecular).


25.

A solution of urea (mol. mass 56g mol-1) boils at 100.18°C at the atmospheric pressure. If kf and kb for water are 1.86 and 0.512 K kg mol-1 respectively, the above solution will freeze at

  • -6.54°C

  • 6.54°C

  • 0.654°C

  • - 0.654°C


D.

- 0.654°C

Tf = kf x molality of solution

   Tb = kb x molality of solution

or TfTb = kfkb

Given that, Tb = T2 - T1 = 100.18 - 100 = 0.18

kf for water = 1.86 K kg mol-1

kb for water = 0.512 K kg mol-1

   Tf0.18 = 1.860.512 Tf               = 1.86 x 0.180.512= 0.6539 ~ 0.654Tf = T1 - T20.654 = 0°C - T2 T2 = - 0.654°C


26.

Names of some compounds are given. Which one is not correct in IUPAC system?

  • CH3 - CH(OH) -CH(CH3) - CH33-methyl-2-butanol

  • CH3 - C C - CH(CH3)4-methyl-2-pentyne

  • CH3 - CH2 - C(CH2) - CH(CH3) -CH32-ethyl-3methyl-but-1-ene

     


D.

Correct IUPAC name is 4-ethyl-3-methyl heptane because substituents are written in alphabetical order. 


27.

Which one of the following forms micelles in aqueous solution above certain concentration?

  • Urea

  • Dodecyl trimethyl ammonium chloride

  • Pyridinium chloride

  • Glucose


B.

Dodecyl trimethyl ammonium chloride

Surfactants detergents form micelles in aqueous solution above to their CMC i.e. Critical Micelle Concentration. Dodecyl trimethyl ammonium chloride is an example of cationic surfactant. Therefore, it shows a following phenomena: 


28.

The aqueous solution containing which one of the following ions will be colourless ? (Atomic no. Sc= 21, Fe= 26, Ti= 22, Mn = 25)

  • Sc3+

  • Fe2+

  • Ti3+

  • Mn2+


A.

Sc3+

21Sc = 1s2, 2s22p6, 3s23p63d1, 4s2

So, Sc3+ = 1s2, 2s22p6, 3s23p6

(It is colourless due to absence of unpaired electrons in d- stubshell)

26Fe = 1s2, 2s22p6, 3s23p63d6

   Fe2+ = 1s2, 2s22p6, 3s23p63d6

(It is coloured due to presence of four unpaired electrons in d-subshell.)

22Ti = 1s2, 2s22p6, 3s23p63d2,4s2

   Ti3+ = 1s2, 2s22p6, 3s2, 3s63d1

(It is coloured due to presence of one unpaired electron in d-subshell).

25Mn = 1s2, 2s22p6, 3s23p63d5, 4s2

 Mn2+ = 1s2, 2s22p6, 3s23p63d5

(It is coloured due to presence of five unpaired electrons in d-subshell).


29.

For a first order reaction A  B, the reaction rate at reactant concentration of 0.01 M is found to be 2.0 x 10-5 mol L-1s-1. The half-life period of the reaction is

  • 220 s

  • 30 s

  • 300 s

  • 347 s


D.

347 s

For first order reaction

B

Given that rate = k x [A]

                Rate =2.0 x 10-5 mol L-1s-1

               [A]    = conc. of A = 0.01M

So, 2.0 x 10-5 = k x 0.01 

                 k   = 2.0 x 10-50.01s-1

                      = 2.0 x 10-3s-1

For first order reaction

T1/=0.693k = 0.6932.0 x 10-3  = 346.5

i.e. 347 s


30.

What is the correct relationship between the pHs of isomolar solutions of sodium oxide (pH1), sodium sulphide (pH2), sodium selenide (pH3) and sodium telluride (pH4)?

  • pH1 > pH2 pH3 > pH4

  • pH1 < pH2 pH3 < pH4

  • pH1 < pH2 pH3 pH4

  • pH1 > pH2 pH3 > pH4


D.

pH1 > pH2 pH3 > pH4

The correct order of pH of isomolar solutions is

pH1 > pH2 > pH3 > pH4

sodium oxide > sodium sulphide > sodium selenide > sodium telluride.

This is because, in aqueous solution they are hydrolysed as follows:

Na2O + 2H22NaOHBase + H2O

Na2S + 2H22NaOHStrong base + H2SWeak acid  

Na2Se + 2H22NaOHStrong base + H2SeWeak acid

Na2Te + 2H22NaOH Strong base + H2TeWeak acid

Therefore, the order of acidic strength is

H2Te > H2Se > H2S > H2

Since, the pH of basic solution is higher than acidic or least basic solution.