Subject

Chemistry

Class

NEET Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

1.

a/ V2 given in van der Waal's equation is for

  • internal pressure

  • inter molecular attraction

  • both (1) and (2)

  • temperature correction


B.

inter molecular attraction

For the explaination of real gas behaviour van der Waal gave an equation. The gas pressure is changed due to intermolecular attraction. By adding aV2 the pressure value is made correct.


2.

C(dia) + O2  CO2H = - 395.4 kJ/ mol

C(gr) + O2 CO2H   = - 393.5 kJ/ mol

C(gr) C(dia)H = ?

  • -3.8

  • -1.9

  • +3.8

  • +1.9


D.

+1.9

C(dia) + O2 CO2 (H = -395.4 kJ/mol)         ...(1)

C(gr) + O2 CO2  (H = -393.5 kJ/mol)         ...(2)

On subtarcting equation (1) from (2), we get,

C(gr) C(dia)H = +1.9 kJ/ mol

Since, H = -393.5 - (-395.4)

                = +1.9 kJ mol-1


3.

Hybridization of 1 and 2 carbon atom in

C1H2C2 = CHare

  • sp, sp

  • sp2, sp2

  • sp2, sp

  • sp3, sp2


C.

sp2, sp

In this compound carbon atoms have following hybridization :

C1H2sp2 = Csp2 =CH2 

Carbon one contains one m bond and carbon two contains two m bonds and hence show sp2 and sp hybridization respectively.


4.

In an isothermal process

  • H = E+PV

  • H = W

  • H = E

  • H =ST


B.

H = W

In an isothermal process, the temperature is constant.

 E = 0Hence, H = PV           (As, H = E + PV)               H = W                ( W = PV)


5.

Which has minimum solubility?

  • Bi2S3

  • Ag2S

  • CoS

  • PbS


B.

Ag2S

Solubilities of salts are as follows:

Bi2S3 = Ksp1085            = 1.6×10-721085 = 1.7 × 10-15Ag2S = Ksop43 = 1.6 × 10-4943            = 3.4 × 10-17PbS  = Ksp = 1.1 ×10-29           = 3.3 × 10-15CoS  = Ksp = 2 × 10-27           = 4.5 × 10-14

Hence, Ag2S is least soluble. Therefore, minimum solubility is of CoS as its force of attraction is greatest.


6.

Ratio of molecular weights of A and B is 425 then ratio of rates of diffusion will be

  • 5 : 1

  • 5 : 2

  • 25 : 3

  • 25 : 4


B.

5 : 2

According to Graham's rule

r1r2 = M2M1r1r2 = 254     = 52

Hence, ratio of diffusion is 5 : 2


7.

The pH of a 10-10M - NaOH solution is

  • 10

  • 7.01

  • 6.99

  • 4


B.

7.01

When the solution is very dilute, the concentration of OHproduced from water cannot be neglected. Hence,

[OH-]  = 10-10 + 10-7 (obtained from water)

          = 10-7 (0.001 + 1)

          = 1.001 × 10-7

 pOH = -log[OH-]

           = -log(1.001 × 10-7)

           = 7 - 0.01 = 6.99

 pH   = 14 - pOH

           = 14 - 6.99

           = 7.01

Therefore, the pH of a 10-10 - NaOH solution is 7.01


8.

Number of orbitals in L energy level

  • 1

  • 2

  • 3

  • 4


D.

4

It is second energy level. Second energy level contains one s-orbital and three p- orbitals. Therefore, total number of orbital in L level is four.


9.

The internal energy of a substance

  • decrease with increase in temperature

  • increase with increase in temperature

  • remains unaffected with change in temperature

  • calculated by E = mc2


B.

increase with increase in temperature

The internal energy of a substance always increases with increase in temperature.


10.

H-bond is strongest in

  • C2H5OH

  • H-F

  • H2O

  • CH3COCH3


B.

H-F

All compound which have hydrogen atom attached to F, O or N atom (i.e. highly electronegative atoms) exhibit H-bonding. In the question H - F exhibit strongest H-bonding because F is the most electronegative element.