Subject

Chemistry

Class

NEET Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

1.

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 chemical species to each other in the balanced equation

2828 Views

2.

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

  • oxygen is more electronegative

942 Views

3.

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 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
1138 Views

4.

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

  • 7.0

4508 Views

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5.

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

  • 60


D.

60

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

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6.

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 

  • 7 g 

1927 Views

7.

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 3d 4s

1078 Views

8.

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

  • high temperatures and high pressures

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9.

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

  • -630 kJ

4764 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

  • H2CO3 and Na2PO4

1214 Views

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