Subject

Chemistry

Class

NEET Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

1.

The Vividh Bharti Station of All India Radio, Delhi, broadcasts on a frequency of 1,368 kHz (Kilohertz). Calculate the wavelength (λ) of the electromagnetic radiation emitted by transmitter. Which part of the electromagnetic spectrum does it belong to?

  • 319.4 m and X-rays

  • 319.4 m and radiowave

  • 219.3 m and microwave

  • 219.3 m and radiowave


D.

219.3 m and radiowave

In the above question,

Frequencyv1368 ×103 Hz

 λ = cv = 3 × 108 m/s1368 × 103 s-1  = 219.3 m

(where, λ = wavelength; c = velocity of light; v = frequency)

The range of wavelength of radiowaves is 100 m to radiation is 10000 m.


2.

0.40 g of helium in a bulb at a temperature of T K had a pressure of p atm. When the bulb was immersed in water bath at temperature 50 K more than the first one, 0.08 g of gas had to be removed to restore the original pressure. T is

  • 500 K

  • 400 K

  • 600 K

  • 200 K


D.

200 K

As pressure and volume remains constant

n1T1 = n2T2

n10.44 = 0.1; T1 = TK

n2 = 0.40 - 0.084 = 0.08; T2 = (T + 50)K

On putting the values,

0.1 x T = 0.08 x (T + 50)

0.1T = 0.08 T = 4

0.1 T - 0.08 T = 4

0.02 T = 4 

T = 200K


3.

Which of the following is correctly arranged in order of increasing weight?

  • 0.0105 equivalent of H2CO2 ·2H2O < 0.625 g of Fe < 0.006 g atom of Ag <6.0 x 1021 atoms of Zn

  • 0.625 g of Fe < 0.0105 equivalent of H2C2O4 ·2H2O < 6.0 x 1021 atoms of Zn < 0.006 g atom of Ag 

  • 0.625 g of Fe < 6.0 x 1021 atoms of Zn < 0.006 g atom of Ag < 0.0105 equivalent of H2C2O4 ·2H2O

  • 0.0105 equivalent of H2C2O4. 2H2O < 0.006 g atom of Ag <6.0 x 1021 atoms of Zn < 0.625 g of Fe


C.

0.625 g of Fe < 6.0 x 1021 atoms of Zn < 0.006 g atom of Ag < 0.0105 equivalent of H2C2O4 ·2H2O

0.625 g Fe

0.0105 eq of H2C2O4. 2H2O = 0.0105 x 63

                                          = 0.66 g (Eq. wt. = 63)

 6 x 1021 atoms of Zn = 6 × 10216.023 × 1023 × 65 = 0.647 g

0.006 g atom of Ag= 0.006 x 108= 0.648 g

Therefore, Increasing order of weight

Fe < Zn < Ag < H2CO3 ·2H2O


4.

What would be the heat released when an aqueous solution containing 0.5 mole of HNOis mixed with 0.3 mole of OH-?  (enthalpy of neutralisation is- 57.1 kJ)

  • 28.5 kJ

  • 17.1 kJ

  • 45.7 kJ

  • 1.7 kJ


B.

17.1 kJ

0.5 mole of HNO3 equivalents to 0.5 mole of H+. But only 0.3 mole of OH- ions are available.

Thus, 0.3 mole of H+ ions will combine with 0.3 mole of OH- ions to form 0.3 mole of H2O and 0.2 mole of H+ ions will remain unreacted.

Heat evolved when 1 mole of H+ ions combine with 1 mole of OH- ions = 57.1 kJ.

Therefore, heat evolved when 0.3 mole of H+ ions combine with 0.3 mole of OH- ions = 0.3x 57.1 = 17.13 kJ.


5.

The charge/size ratio of a cation determines its polarising power. Which one of the following sequences represents the increasing order of the polarising power of the cationic species, K+, Ca2+, Mg2+, Be2+?

  • K< Ca2+ < Mg2+ < Be2+

  • Be2+ < K2+ < Ca2+ < Mg2+

  • Mg2+< Be2+ < K+ < Ca2+

  • Ca2+ < Mg2+ < Be+ < K+


A.

K< Ca2+ < Mg2+ < Be2+

High charge and small size of the cations increases polarisation. As the size of the given cations decreases as

K> Ca2+ > Mg2+ > Be2+

Thus, polarising power increases as

K+ < Ca2+ < Mg2+ < Be2+


6.

In an amino acid, the carboxyl group ionises at pKa = 2.34 and ammonium ion at pKa2 = 9.60. The isoelectric point of the amino acid is at pH

  • 4.32

  • 3.34

  • 9.46

  • 5.97


D.

5.97

Isoelectric point, pH = pKa1 + pKa22 

                              = 234 + 9.602

                               = 5.97


7.

In an atom, an electron is moving with a speed of 600 m/s with an accuracy of 0.005%. Certainty with which the position of the electron can be located is

(Given, h= 6.6 x 10-34 kg m2 s-1, mass of electron em = 9.1 x 10-31 kg)

  • 215 x 10-3 m

  • 2.78 x 10-3 m

  • 1.92 x 10-3 m

  • 3.24 x 10-3 m


C.

1.92 x 10-3 m

Given that velocity of electron, v = 600 m/s

Accuracy of velocity = 0.005%

 v = 600 × 0.005100 = 0.03

According to Heisenberg's uncertainty principle

x . mv  h4π

x = 6.6 × 10-344 × 3.14 × 9.1 × 10-31 ×0.03

x =1.92 × 10-3 m


8.

The number average molar mass and mass average molar mass of a polymer are respectively 30,000 and 40,000. The
polydispersity index (PDI) of the polymer is 

  • -1

  • 0

  • > 1

  • < 1


C.

> 1

Polydispersity Index or PDI = Mass average molar massNumber average molar mass = MwMn

PDI = 4000030000 = 1.3 > 1

Therefore, polydispersity index or pDI of the polymer is > 1.


9.

For preparing 3.00 L of 1M NaOH by mixing portions of two stock solutions (A and B) of 2.50 M NaOH and 0.40 M NaOH respectively. Find out the amount of B stock solution (in L) added.

  • 8.57 L

  • 2.14 L

  • 1.28 L

  • 7.51 L


B.

2.14 L

Moles needed = 3.00 L × 1.00 M = 3 mol.

Let's assume, x L of 2.50 M NaOH is added. Then (3 - x) L of 0.40 M NaOH is added (Given). ..... (i)

The number of moles of solute from the more concentrated solution is 2.50, that becomes less concentrated solution is (0.40) (3.00 - x).

Total number of moles = 3.00

 2.50x + 0.40 (3.00 - x) = 3.00

    2.50x + 1.2 - 0.40x = 3.00

    2.10x + 1.2 = 3.00

    2.10x = 1.8

     = 0.857 L of 2.50 M NaOH  ... (ii)

Putting value of x from (ii) in (i), is

3 - 0.857 = 2.14 L of 0.40 M NaOH

Therefore, the amount of B stock solution to be added is 2.14 L.


10.

What percentage of β-D-(+-glucopyranose is found at equilibrium in the aqueous solution?

  • 64%

  • 36%

  • 100%

  • 50%


A.

64%

Ordinary glucose is α-glucose, with a fresh aqueous solution has specific rotation, [α]D = + 110°. On keeping the solution for sometimes, α-glucose slowly changes into an equilibrium mixture of α-glucose (36%) and β-glucose (64%) and the mixture has specific rotation+ 52.5°.