Subject

Chemistry

Class

NEET Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

11.

The value of ΔH for the reaction

X2 (g) + 4Y2 (g) ⇌ 2XY4 (g) is less than zero. Formation of XY4 (g) will be favoured at

  • Low pressure and low temperature

  • high temperature and low pressure

  • high pressure and low temperature

  • high pressure and low temperature


C.

high pressure and low temperature

X2 (g) + 4Y2 (g) ⇌ 2XY4 (g);


ΔH < 0 and Δn < 0

Therefore, the forward reaction is favoured at high pressure and low temperature. (According to Le- Chatelier's principle)
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12.

Mole fraction of the solute in a 1.00 molal aqueous solution is 

  • 0.0177

  • 0.0344

  • 1.7700

  • 1.7700


A.

0.0177

1.00 molal aqueous solution = 1.0 mole in 1000 g water nsolute = 1; Wsolvent = 1000g

nsolvent = 1000/18 = 55.56

Xsolute = 1/ 1+ 55.56 = 0.0177

449 Views

13.

By what factor does the average velocity of a gaseous molecule increase when the temperature (in kelvin) is doubled?

  • 2.8

  • 4.0

  • 1.4

  • 1.4


C.

1.4

Average space velocity space equals space square root of fraction numerator 8 RT over denominator πM end fraction end root
therefore space straight V subscript av space proportional to space square root of straight T

or space fraction numerator left parenthesis straight V subscript av right parenthesis subscript 2 over denominator left parenthesis straight V subscript av right parenthesis subscript 1 end fraction space equals space square root of fraction numerator 2 straight T over denominator straight T end fraction end root space equals space 1.4
644 Views

14.

If n =6, the correct sequence for filling of electrons will be

  • ns → (n-1) d → ( n - 2)f → np

  • ns - (n-2)f → np → (n-1)d

  • ns - np → (n-1)d → (n-2)f

  • ns - np → (n-1)d → (n-2)f


D.

ns - np → (n-1)d → (n-2)f

6s → 4f → 5d → 6p for n = 6

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15.

The total number of atomic orbitals in fourth energy level of an atom is 

  • 16

  • 32

  • 4

  • 8


A.

16

 

 

Number of atomic orbitals in an orbit is given by,

= n2 = 42 = 16

The number of atomic orbitals in fourth energy level  is = 16

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16.

For the reaction N2 (g) + O2 (g) ⇌  2NO (g), the equilibrium constant is K1. The equilibrium constant is K2 for the reaction 2NO (g) + O2(g) ⇌ 2NO2 (g). What is K for the reaction NO subscript 2 space left parenthesis straight g right parenthesis space rightwards harpoon over leftwards harpoon 1 half straight N subscript 2 space left parenthesis straight g right parenthesis space plus straight O subscript 2 space left parenthesis straight g right parenthesis ? space space

  • 1/(4K1K2)

  • [1/K1K2]1/2

  • 1/(K1K2)

  • 1/(K1K2)


B.

[1/K1K2]1/2

N2 (g) + O2 (g) ⇌  2NO (g) ; K1 (i)
2NO (g) + O2 (g) ⇌ 2NO2 (g) ; K2 (ii)
______________________________
N2 (g) + O2(g)  ⇌ 2NO2 (g) ; K = K1 x K2

therefore, For NO2 (g) = N2/2 (g)  +O2 (g);

K = [1/K1K2]1/2

1715 Views

17.

The energies E1 and E2 of two radiations are 25 eV respectively. The relation between their wavelength i.e., λ1 and λ2 will be

  • λ1 = 2λ2

  • λ1 = 4λ2

  • λ1 = λ2/2

  • λ1 = λ2/2


A.

λ1 = 2λ2

E1 = 25eV. E2 = 50 eV

E1 = hc/λ1  and E2 = hc/λ2

fraction numerator straight E subscript 1 over denominator straight E subscript 2 space end fraction space equals space lambda subscript 1 over lambda subscript 2
p u t t i n g space t h e space g i v e n space v a l u e
25 over 50 space equals space lambda subscript 1 over lambda subscript 2
rightwards double arrow space lambda subscript 1 space equals space 2 lambda subscript 2

1680 Views

18.

In Duma's method of estimation of nitrogen 0.35 g of an organic compound gave 55 mL of nitrogen collected at 300 K temperature and 715 mm pressure. The percentage composition of nitrogen in the compound would be

(Aqueous tension at 300 k = 15 mm)

  • 16.45

  • 17.45

  • 14.45

  • 14.45


A.

16.45

Gas equation,

fraction numerator straight p subscript 1 straight V subscript 1 over denominator straight T subscript 1 space end fraction space equals space fraction numerator straight p subscript 2 straight V subscript 2 over denominator straight T subscript 2 end fraction
Where comma space straight p subscript 2 space equals space pressure space of space straight N subscript 2 space at space STP space equals space 760 space mm
straight T subscript 2 space equals space Temperature space of space straight N subscript 2 space straight t space STP space equals space 273 straight K
straight V subscript 2 space equals space ?
Volume space of space straight N subscript 2 space at space STP space left parenthesis by space Gas space equation right parenthesis
open parentheses fraction numerator straight rho space minus space straight rho subscript 1 over denominator straight t plus 273 end fraction close parentheses straight V subscript 1 space straight x space 273 over 160 space equals space straight V subscript 2
Where space straight p subscript 1 space equals space straight rho space minus space straight rho subscript 1
straight rho space equals space 715 space mm space left parenthesis pressure space at space which space straight N subscript 2 space collected right parenthesis
straight rho subscript 1 space equals space aqueous space tension space of space water space equals 15 mm
straight T subscript 1 space equals space straight t space plus space 273 space equals space 300 space straight k space
straight V subscript 1 space equals space 55 space mL space space equals space volume space of space moist space nitrogen space in space nitrometer
therefore space straight V subscript 2 space equals space fraction numerator left parenthesis 715 minus 15 right parenthesis space straight x space 55 over denominator 300 end fraction space straight x space 273 over 760 space equals space 40.098 space mL
percent sign space of space nitrogen space in space given space compound

equals space 28 over 22400 space straight x space straight V subscript 2 over straight W space straight x space 100 space

equals space 2 over 22400 space straight x space fraction numerator 46.09 over denominator 0.35 end fraction space straight x space 100 thin space

space equals space 16.45 percent sign

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19.

Which one of the following statements is not true?

  • Concentration of DO below 6 ppm is good for the growth of fish

  • Clean water would have a BOD value of less than 5 ppm

  • Oxides of sulphur, nitrogen and carbon are the most widespread air pollutant

  • Oxides of sulphur, nitrogen and carbon are the most widespread air pollutant


A.

Concentration of DO below 6 ppm is good for the growth of fish

The growth of fish is inhibited if the concentration of DO is below 6 ppm.

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20.

Which of the following has the minimum bond length?

  • O2-

  • O22-

  • O2

  • O2


D.

O2

Bond space order space equals fraction numerator space straight n subscript straight b space minus straight n subscript straight a over denominator 2 end fraction
therefore space the space bond space order space of space given space molecule
Bond space order space of space straight O subscript 2 superscript space plus end superscript space equals space fraction numerator 10 minus 5 over denominator 2 end fraction space equals space 2.5

Bond space order space of space straight O subscript 2 superscript space minus end superscript space equals space fraction numerator 10 minus 7 over denominator 2 end fraction space equals space 1.5

Bond space order space of space straight O subscript 2 superscript space 2 minus end superscript space equals space fraction numerator 10 minus 8 over denominator 2 end fraction space equals space 1

Bond space order space of space straight O subscript 2 superscript space space equals space fraction numerator 10 minus 6 over denominator 2 end fraction space equals space 2
therefore, maximum bond order = minimum bond length.
Hence bond length is minimum for O2+
601 Views