Subject

Chemistry

Class

NEET Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

1.

Decreasing order of stability of straight O subscript 2 comma space straight O subscript 2 superscript minus comma space straight O subscript 2 superscript plus comma space and space straight O subscript 2 superscript 2 minus end superscript is

  • straight O subscript 2 superscript plus space greater than space straight O subscript 2 space greater than space straight O subscript 2 superscript minus space greater than space straight O subscript 2 superscript 2 minus end superscript
  • straight O subscript 2 superscript 2 minus end superscript space greater than space straight O subscript 2 superscript minus space greater than space straight O subscript 2 space greater than space straight O subscript 2 superscript plus
  • straight O subscript 2 space greater than straight O subscript 2 superscript plus space greater than space straight O subscript 2 superscript 2 minus end superscript space greater than thin space straight O subscript 2 superscript minus
  • straight O subscript 2 superscript minus space greater than thin space straight O subscript 2 superscript 2 minus end superscript space greater than thin space straight O subscript 2 superscript plus space greater than thin space straight O subscript 2

A.

straight O subscript 2 superscript plus space greater than space straight O subscript 2 space greater than space straight O subscript 2 superscript minus space greater than space straight O subscript 2 superscript 2 minus end superscript

order of stability directly proportional to the bond order

therefore, the order of the stability of given species,

stack space space with Bond space order below space space space stack straight O subscript 2 superscript plus with 2.5 below space greater than stack space straight O subscript 2 space with 2 below greater than stack space straight O subscript 2 superscript minus with 1.5 below space greater than space stack straight O subscript 2 superscript 2 minus end superscript with 1 below

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2.

20.0 g of a magnesium carbonate sample decomposes on heating to give carbon dioxide and 8.0 g magnesium oxide. What will be the percentage purity of magnesium carbonate in the sample? (Atomic weight of Mg =24)

  • 75

  • 96

  • 60

  • 84


D.

84

In the given problem we have a practical yield of MgO. For calculation of percentage yield of MgO. we need a therortical yield of MgO. For this, we shall use mole concept.

MgCO3(s) → MgO (s) + CO2(g)  .. (i)


Moles space of space MgCO subscript 3 subscript space space end subscript equals space fraction numerator weight space in space gram over denominator Molecular space weight end fraction

space equals space 20 over 84 space equals space 0.238 space mol

From space eq space left parenthesis straight i right parenthesis space
1 space mole space of space MgCO subscript 3 space gives space equals space 1 space mol space MgO
there space 0.238 space mole space MgCO subscript 3 space will space give space equals space 0.238 space mol space MgO
space equals space 0.238 space straight x space 40 space straight g space equals space 9.52 space straight g space MgO
Now comma space practical space yield space of space MgO space equals space 8 space straight g space
therefore comma space percent sign purity space equals space fraction numerator 8 over denominator 9.52 end fraction space straight x space 100 space equals space 84 percent sign

bold Alternate bold space bold Method
stack MgCO subscript 3 with 84 space straight g below space rightwards arrow space stack MgO space with 40 space straight g below plus space CO subscript 2

therefore comma space 8 space straight g space MgO space will space form space 84 over 5 straight g
therefore space comma percent sign space purity space equals space 84 over 5 space straight x space 100 over 20 space equals space 84 percent sign

4442 Views

3.

The formation of the oxide ion O2- (g), from oxygen atom requires first an exothermic and then an endothermic step as shown below,

straight O space left parenthesis straight g right parenthesis space plus space straight e to the power of minus space rightwards arrow space straight O to the power of minus left parenthesis straight g right parenthesis semicolon space increment subscript straight f straight H to the power of straight o space equals space minus 141 space kJ space mol to the power of negative 1 end exponent
straight O to the power of minus space left parenthesis straight g right parenthesis space plus straight e to the power of minus space rightwards arrow space straight O to the power of 2 minus end exponent space left parenthesis straight g right parenthesis semicolon space increment subscript straight f straight H to the power of straight o space equals plus 780 space kJ space mol to the power of negative 1 end exponent
Thus, the process of formation of O2- in the gas phase is unfavourable even though O2- is isoelectronic with neon. It is due to the fact that

  • electron repulsion outweighs the stability gained by achieving a noble gas configuration

  • O- ion has comparatively smaller size that oxygen atom

  • oxygen is more electronegative

  • the addition of electron in oxygen result in the large size of the ion


A.

electron repulsion outweighs the stability gained by achieving a noble gas configuration

Electron repulsion predominates over the stability gained by achieving noble gas configuration. Hence, the formation of O2- in the gas phase is unfavourable.

942 Views

4.

What is the mass of precipitate formed when 50 mL of 16.9% solution of AgNO3 is mixed with 50 mL of 5.8% NaCl solution?

(Ag = 107.8, N = 14, O = 16, Na = 23, Cl = 35.5)


  • 28 g

  • 3.5 g 

  • 7 g 

  • 14 g 


C.

7 g 

For the calculation of the mass of AgCl precipitate, we find the mass of AgNO3 and NaCl in equal volume with the help of mole concept. 16.9% solution of AgNO3 means 16.9 g AgNO3 is present in 100 mL solution.
therefore, 8.45 g AgNO3 will present in 50 mL solution similarly,
5.8 g NaCl is present in 100 ml solution

stack space space space space space space space space space space space space space space space space space space space space space space space space with Initial space mole below space space stack space AgNO subscript 3 with fraction numerator 8.45 over denominator 169.8 end fraction below space plus space NaCl with fraction numerator 2.9 over denominator 58.5 end fraction space space space space space space below space rightwards arrow space AgCl with 0 below space plus space stack NaNO subscript 3 with 0 below
space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 0.049 space space space space space equals 0.049
After space reaction space space space space space space 0 space space space space space space space space space space space space space space space space space space 0 space space space space space space space space space 0.049 space space space space space space space space space 0.049
therefore comma space Mass space of space AgCl space precipitated
space equals space 0.049 space straight x space 143.5 space equals 7 space straight g

1927 Views

5.

The heat of combustion of carbon to CO2 is -395.5 kJ/mol. The heat released upon the formation of 35.2 g CO2 from carbon and oxygen gas is 

  • -315 kJ

  • +315 kJ

  • -630 kJ

  • -318 kJ


A.

-315 kJ

Given comma space straight C space left parenthesis straight s right parenthesis space plus space straight O subscript 2 space left parenthesis straight g right parenthesis space rightwards arrow space CO subscript 2 space left parenthesis straight g right parenthesis semicolon
increment subscript straight f straight H space equals space minus 393.5 space kJ space mol to the power of negative 1 end exponent
therefore comma space Heat space released space on space formation space of space 44 space straight g space or space 1 space mole
CO subscript 2 space equals space minus 395 space kJ space mol
Therefore comma
Heat space released space on space formation space of space 35.2 space straight g space of space CO subscript 2
space equals space fraction numerator negative 393.5 space kJ space mol to the power of negative 1 end exponent over denominator 44 space straight g end fraction space straight x space 35.2 space straight g space equals space minus space 315 space kJ space mol to the power of negative 1 end exponent
4764 Views

6.

What is the pH of the resulting solution when equal volumes of 0.1 M NaOH and 0.01 M HCl are mixed?

  • 12.65

  • 2.0

  • 7.0

  • 1.04


A.

12.65

When equal volumes of acid and base are mixed then resulting solution become alkaline if the concentration of base is taken high.

Let normality  of the solution after mixing 0.1 M
NaOH and 0.01 M HCl is N.
therefore,   N1V1 - N2V2 = NV

or   0.1 x 1 - 0.01 x 1 = N x 2

Since normality, of NaOH, is more than that of HCl. Hence, the resulting solution is alkaline.

Or

[OH-] = N = 0.09/2 = 0.045 N

Or pOH = - log (0.045) = 1.35

therefore, pH = 14-pOH

= 14 - 1.35 = 12.65

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7.

If Avogadro number NA, changed from 6.022 x 1023 mol-1 this would change

  • the definition of mass in units of grams

  • the mass of one mole of carbon

  • the ratio of chemical species to each other in the balanced equation

  • the ratio of elements to each other in a compound


B.

the mass of one mole of carbon

If Avogadro number NA, is changed from 6.022 x 1023 mol-1 to 6.022 x 1020 mol-1 this would change the mass of one mole of carbon.

therefore,

1 mole of carbon has mass = 12 g

or 6.022 x 1023 atoms of carbon have mass = 12 g 

therefore, 6.022 x 1020 atoms of carbon have mass


fraction numerator 12 over denominator 6.022 space straight x space 10 to the power of 23 end fraction space x space 6.022 space x space 10 to the power of 20 space equals space 0.012 g

2828 Views

8.

Which is the correct order of increasing energy of the listed orbitals in the atom of titanium?

  • 3s 4s 3p 3d

  • 4s 3s 3p 3d

  • 3s 3p 3d 4s

  • 3s 3p 4s 3d


C.

3s 3p 3d 4s

According to Aufbau rule

3s < 3p < 3d< 4s

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9.

A gas such as carbon monoxide would be most likely to obey the ideal gas law at

  • high temperatures and low pressures

  • low temperatures and high pressures

  • high temperatures and high pressures

  • low temperatures and low pressures


A.

high temperatures and low pressures

Real gases show ideal gas behaviour at high temperatures and low pressures.
 
PV = nRT

2564 Views

10.

Which one of the following pairs of the solution is not an acidic buffer?

  • HClO4 and NaClO4

  • CH3COOH and CH3COONa

  • H2CO3 and Na2PO4

  • H2PO4 and Na3PO4


A.

HClO4 and NaClO4

A buffer solution is one which resists changes in pH when small quantities of an acid or an alkali are added to it.

HClO4 and NaClO4 are not an acidic buffer because strong acid with its salt cannot form buffer solution.
1214 Views