Subject

Chemistry

Class

NEET Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

1.

which of the following orders of ionic radii is correctly represented?

  • H- > H >H+

  • Na+ >F- >O2-

  • F- > O2->Na+

  • Al3+> Mg2+>N3-


A.

H- > H >H+

It is known that radius of the cation is always smaller than that of the neutral atom due to decrease in the number of orbits.Whereas, the radius of the anion is always greater than a cation due to a decrease in effective nuclear charge. Hence, the correct order is 

H- > H >H+

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2.

For the reaction X2O4 (l) -->  2XO2 (g) ΔU = 2.1 kcal, ΔS = 20 cal K-1 at 300 K hence ΔG is

  • 2.7 kcal

  • -2.7kcal

  • 9.3 kcal 

  • -9.3 kcal


B.

-2.7kcal

The change in Gibbs free energy is given by

ΔG = ΔH-TΔS

Where, ΔH = enthalpy of the reaction

ΔS = entropy of reaction
Thus, in order to determine ΔG, the values of ΔH must be known. The value of ΔH can be calculated by the equation
ΔH = ΔU + ΔngRT
Where (ΔU) = change in internal energy

Δng = (number of moles of gaseous products)-(number of moles of gaseous reactant) = 2-0 =2
R = gas constant = 2 cal
But, ΔH= Δu +  ΔngRT 
Δu  =2.1 kcal = 2.1 x 103 cal

[1kcal = 103 cal]
therefore,
ΔH = (2.1 x 103) +(2x2x300) =3300 cal
Hence, ΔG = ΔH-TΔS
 ΔG = (3300)-(300 x20)
 ΔG =-2700 cal
 ΔG =-2.7 Kcal

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3.

Which of the following salts will give highest pH in water?

  • KCl

  • NaCl

  • Na2CO3

  • CuSO4


C.

Na2CO3

The highest pH refers to the basic solution containing OH- ions. Therefore, the basic salt releasing OH- ions on hydrolysis will give highest pH in water.
Only the salt of a strong base and weak acid would release OH-ion on hydrolysis. Among the given salts, Na2CO3 corresponds to the basic salt as it is formed by the neutralisation of NaOH [strong base] and H2CO3 [weak acid].

CO subscript 3 superscript 2 minus end superscript space plus straight H subscript 2 straight O space rightwards harpoon over leftwards harpoon HCO subscript 3 superscript minus space plus OH to the power of minus

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4.

For the reversible reaction,

straight N subscript 12 space left parenthesis straight g right parenthesis space plus 3 straight H subscript 2 left parenthesis straight g right parenthesis space rightwards harpoon over leftwards harpoon 2 NH subscript 3 space left parenthesis straight g right parenthesis space plus Heat
the equilibrium shifts in the forward direction

  • by increasing the concentration of NH3 (g)

  • by decreasing the pressure

  • by decreasing the concentrations of N2 (g)  and H2(g)

  • by increasing pressure and decreasing the temperature


D.

by increasing pressure and decreasing the temperature

Any change in the concentration, pressure and temperature of the reaction results in a change in the direction of equilibrium.This change in the direction of equilibrium in governed by Le-Chatelier's principle. According to Le-Chatelier's principle, the equilibrium shifts in the opposite direction to undo the change.

straight N subscript 12 space left parenthesis straight g right parenthesis space plus 3 straight H subscript 2 left parenthesis straight g right parenthesis space rightwards harpoon over leftwards harpoon 2 NH subscript 3 space left parenthesis straight g right parenthesis
Increasing pressure and decreasing temperature
On increasing pressure, equilibrium shifts in the forward direction where the number of moles decreases while on decreasing temperature. it will move in a forward direction where temperature increases.

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5.

Equal masses of H2, O2 and methane have been taken in a container of volume V at temperature 270 C in identical conditions. The ratio of the volumes of gases H2 : O2: CH4 would be 

  • 8:16:1

  • 16:8:1

  • 16:1:2

  • 8:1:2


C.

16:1:2

According to Avogadro's hypothesis,
Volume of a gas (V) straight alpha number of moles (n)
Therefore, the ratio of the volumes of gases can be determined in terms of their moles. The ratio of volumes of H2:O2: methane (CH4) is given by

straight V subscript straight H subscript 2 end subscript space colon space straight V subscript straight O subscript 2 end subscript space colon space straight V subscript CH subscript 4 end subscript space equals space straight n subscript straight H subscript 12 end subscript space colon straight n subscript straight O subscript 2 end subscript space colon space straight n subscript CH subscript 4 end subscript

rightwards double arrow space straight V subscript straight H subscript 2 end subscript space colon space straight V subscript straight O subscript 2 end subscript space colon space straight V subscript CH subscript 4 end subscript space equals space straight m subscript straight H subscript 2 end subscript over straight M subscript straight H subscript 2 end subscript colon space straight m subscript straight O subscript 2 end subscript over straight M subscript straight O subscript 2 end subscript space colon straight m subscript CH subscript 4 end subscript over straight M subscript CH subscript 4 end subscript
But space straight m subscript straight H subscript 2 end subscript space equals space straight m subscript straight O subscript 2 end subscript space equals space straight m subscript CH subscript 4 end subscript space equals space straight m space open square brackets therefore space straight n space equals space fraction numerator mass over denominator molar space massd end fraction close square brackets
Thus comma space space straight V subscript straight H subscript 2 end subscript space colon space straight V subscript straight O subscript 2 end subscript space colon space straight V subscript CH subscript 4 end subscript space space equals space straight m over 2 space equals space straight m over 1 space equals straight m over 16 space equals space 16 colon space 1 colon 2

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6.

What is the maximum number of orbitals that can be identified with the following quantum numbers?

n=3, l =1, m1 = 0

  • 1

  • 2

  • 3

  • 4


A.

1

The value of n=3 and l =1 suggest that it is a 3p orbital while the value of m1 = 0 [magnetic quantum number] shows that the given 3p orbital is 3pz in nature.
Hence, the maximum number of orbitals identified by the given quantum number is only 1, i.e. 3pz.

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7.

Calculate the energy in joule corresponding to light of wavelength 45mm: (Planck's constant h=6.63 x10-34)Js; speed of light c= 3 x 108 ms-1)

  • 6.67 x 1015

  • 6.67 x 1011

  • 4.42 x 10-15

  • 4.42 x 10-18


D.

4.42 x 10-18

The wavelength of light is related to its energy by the equation,

straight E space equals hc over straight lambda
Given comma
straight lambda space equals space 45 space mm space equals space 45 space straight x space 10 to the power of negative 9 end exponent space straight m space
left square bracket therefore space 1 space nm space equals space 10 to the power of negative 9 end exponent space straight m space right square bracket

Hence comma space straight E equals fraction numerator 6.63 space straight x space 10 to the power of negative 34 end exponent space js space straight x space 3 space straight x space 10 to the power of 8 space ms to the power of negative 1 end exponent over denominator 45 space straight x space 10 to the power of negative 9 end exponent space straight m end fraction
space equals space 4.42 space straight x space 10 to the power of negative 18 end exponent space straight j

Hence comma space the space energy space corresponds space to space light space of space wavelenght space 45 space nm space is space 4.42 space straight x space 10 to the power of negative 18 space end exponent space straight j

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8.

When 22.4 L of H2 (g) is mixed with 11.2 L of Cl2 (g), each of at STP, the moles of HCl (g) formed is equal to

  • 1 mole of HCl (g)

  • 2 moles of HCl (g)

  • 0.5 mole of (g) 

  • 1.5 mole of HCl(g)


A.

1 mole of HCl (g)

The given problem is related to the concept of the stoichiometry of chemical equations. Thus, we have to convert the given volumes into their moles and then identify the limiting reagent [possessing minimum number of moles and completely used up in the reaction]
The limiting reagent gives the moles of product formed in the reaction.

space space space space space space space space space space space space space space space space space space space space straight H subscript 2 space left parenthesis straight g right parenthesis space plus space space Cl subscript 2 space left parenthesis straight g right parenthesis space rightwards arrow 2 HCl space left parenthesis straight g right parenthesis
initial space vol. space space 22.4 space straight L space space space space space space space space 11.2 space straight L space space space space space space 2 space mol
therefore space 22.4 space straight L space volume space at space STP space is space occupied space by
Cl subscript 2 space equals space 1 space mole
therefore space 11.2 space straight L space volume space will space be space occupied space by comma
Cl subscript 2 space equals space fraction numerator 1 space straight x space 11.2 over denominator 22.4 end fraction space mole
space equals space 0.5 space mol
Thus comma stack space straight H subscript 2 space with 1 mole below left parenthesis straight g right parenthesis space plus stack Cl subscript 2 with 0.5 below space left parenthesis straight g right parenthesis space rightwards arrow 2 HCl space left parenthesis straight g right parenthesis
Since comma Cl subscript 2 space possesses space minimum space number space of space moles comma
thus space it space is space the space limiting space reagent.
As space per space equation comma
1 space mol space Cl subscript 2 space identical to space 2 space mol space HCl
therefore space 0.5 space mol space Cl subscript 2 space equals space 2 space straight x 0.5 space mol space HCl
Hence comma space 1.0 space mole space of space HCl space left parenthesis straight g right parenthesis space is space priduced space by space 0.5 space mole space of space Cl subscript 2.

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9.

For a given exothermic reaction Kp and Kp' are the equilibrium constant at temperatures T1 and T2 respectively. Assuming that heat of reaction si constant in temperature range between T1 and T2 it is readily observed that

  • Kp> Kp'

  • Kp< Kp'

  • Kp = Kp'

  • Kp =1/ Kp'


A.

Kp> Kp'

The equilibrium constant at two different temperatures for a thermodynamic process is given by

log straight K subscript 2 over straight K subscript 1 space equals space fraction numerator increment straight H over denominator 2.303 space straight R end fraction space open square brackets 1 over straight T subscript 1 minus 1 over straight T subscript 2 close square brackets
here space straight K subscript 1 space amd space straight K subscript 2 space are space replaced space by space straight K subscript straight p space and space straight K subscript straight p superscript apostrophe.
Therefore comma space log space fraction numerator straight K subscript straight p superscript apostrophe over denominator straight K subscript straight p end fraction space equals space fraction numerator increment straight H to the power of degree over denominator 2.303 straight R end fraction open square brackets 1 over straight T subscript 1 minus 1 over straight T subscript 2 close square brackets
For space exothermic space reaction comma space straight T subscript 2 space greater than straight T subscript 1 space and space increment straight H space equals space minus ve space
straight K subscript straight p greater than space straight K subscript straight p superscript apostrophe

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10.

Which of the following statement is correct for the spontaneous absorption of a gas?

  • ΔS is negative and therefore, ΔH should be highly positive
  • ΔS is negative and therefore, ΔH should be highly negative
  • ΔS is positive and therefore, ΔH should be negative
  • ΔS is positive and therefore, ΔH should also be highly positive

B.

ΔS is negative and therefore, ΔH should be highly negative
ΔS [change in entropy] and ΔH [change in enthalpy] are related by the equation

ΔG = ΔH-TΔS
[here, ΔG = change in Gibbs free energy]
For adsorption of a gas, ΔS is negative because randomness decreases. Thus, in order to make ΔG negative [for spontaneous reaction], ΔH must be highly negative. Hence, for the adsorption of gas, if ΔS is negative, therefore,ΔH should be highly negative. 
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