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NEET Chemistry Solved Question Paper 2016

Multiple Choice Questions

11.

A 100⁰C the vapour pressure of a solution of 6.5 g of a solute in 100 g water is 732 mm. If Kb = 0.52, the boiling point of this solution will be

100⁰C
102⁰C
103⁰C
103⁰C

103⁰C

According space to space Raoult apostrophe straight s space law space of space paritial space pressure. fraction numerator straight p subscript straight A superscript 0 minus straight p subscript straight s over denominator straight p subscript straight s end fraction space equals straight n subscript straight B over straight n subscript straight A rightwards double arrow space fraction numerator 760 minus 732 over denominator 732 end fraction space equals space fraction numerator straight W subscript straight B space straight x space straight M subscript straight A over denominator straight M subscript straight B space xW subscript straight A end fraction rightwards double arrow space 28 over 732 space equals space fraction numerator 6.5 space straight x space 18 over denominator straight M subscript straight B space straight x space 100 end fraction rightwards double arrow straight M subscript straight B space equals space 30.6 therefore space increment straight T subscript straight b space equals 0.52 space straight x fraction numerator 6.5 space straight x 1000 over denominator 30.6 space space straight x space 100 end fraction space equals 1.10 therefore space Boiling space point space equals 100 plus 1.10 equals 101.1 degree straight C space almost equal to space 101 degree straight C

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12.

The correct statement regarding the comparison of staggered and eclipsed conformations of ethane is

The eclipsed conformation of ethane is more stable than staggered conformation because eclipsed conformation has no torsional strain
The eclipsed conformation of ethane is more stable than staggered conformation even though the eclipsed conformation has torsional strain
The staggered conformation of ethane is more stable than eclipsed conformation because staggered conformation has no torsional strain
The staggered conformation of ethane is more stable than eclipsed conformation because staggered conformation has no torsional strain

The staggered conformation of ethane is more stable than eclipsed conformation because staggered conformation has no torsional strain

Due to the absence of torsional strain staggered conformation of ethane is more stable than eclipsed conformation of it.

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13.

Which of the following statement about hydrogen is incorrect?

Hydrogen never acts as cation in ionic salts
Hydronium ion, H₃O⁺ exist freely in solution
Dihydrogen does not act as a reducing agent
Dihydrogen does not act as a reducing agent

Dihydrogen does not act as a reducing agent

We can have both answers (C, D)
For ionic salts, hydrogen never behaves as cation, but behaves as anion (H^-)
H₃O⁺ exist freely in solution
Dihydrogen acts as a reducing agent.
Hydrogen has three isotopes. Protium, Deuterium and Tritium. Protium is the most common isotopes of hydrogen with an abundance of 99.98%.

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14.

Which is the correct statement for the given acids?

Phosphinic acid is a monoprotic acid while phosphonic acid is a diprotic acid.
phosphinic acid is a diprotic acid while phosphonic acid is a monoprotic acid
Both are triprotic acids
Both are triprotic acids

Phosphinic acid is a monoprotic acid while phosphonic acid is a diprotic acid.

Phosphinic acid

Due to the presence of one replaceable proton in phosphinic acid, it is monoprotic acid. and due to the presence of two replaceable proton in phosphinic acid, it is diprotic acid.

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15.

Which one of the following statements is correct when SO₂ is passed through acidified K₂Cr₂O₇ solution?

The solution is decolourized
SO₂ is reduced
Green Cr₂(SO₄)₃ is formed
The solution turns blue

Green Cr₂(SO₄)₃ is formed

When SO2 is passed through the acidified K₂Cr₂O₇ solution, the orange colour of potassium dichromate solution turns to clear green due to the formation of chromium sulphate. In this reaction, the oxidation state of Cr changes from +6 to +3.

$\underset{OS of Cr = + 6}{K_{2} {Cr}_{2} O_{7}} + H_{2} {SO}_{4} + 3 {SO}_{2} \to K_{2} {SO}_{4} + \underset{Os of Cr = + 3 green}{{Cr}_{2} ({SO}_{4})_{3}} + H_{2} O$

The appearance of green colour is due to the reduction of chromium metal.

16.

Match items of Columns I with the items of Columns II and assign the correct code.

Column I	Column II
A.Cyanide process	1.Ultrapure Ge
B. Froth floatation process	2.Dressing of ZnS
C.Electrolytic reduction	3.Extraction of Al
D.Zone refining	4.Extraction of Au
	5.Purification of Ni

A

B

C

D

2 3 1 5
A

B

C

D

1 2 3 4
A

B

C

D

3 4 5 1
A

B

C

D

3 4 5 1

3 4 5 1

Cyanide process is used for extracting Au (gold) from low-grade ore by converting the Au to the water soluble coordination complex.

Froth flotation process is used for the dressing of sulphide ore such ZnS.

Electrolytic reduction is used for the extraction of Al. The extraction of aluminium is carried out in a steel tank lined inside with graphite. Here graphite used serves as a cathode.

Zone refining is used for ultra pure Ge element. An ingot of Ge is first purified by zone refining. Then a small amount of antimony is placed in the molten zone which is passed through the pure Ge with the proper choice of rate of heating and other variables.

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17.
MY and NY₃, two nearly insoluble salts. have the same K_sp values of 6.2 x 10^-13 at room temperature. Which statement would be true in regard to MY and NY₃ ?

The molar solubility of MY in water is less than that of NY₃

The salts MY and NY₃ are more soluble in 0.5 M KY than in pure water

The addition of the Salt of KY to the solution of MY and NY₃ will have no effect on their solubilities.

The addition of the Salt of KY to the solution of MY and NY₃ will have no effect on their solubilities.

The molar solubility of MY in water is less than that of NY₃

For MY,

$MY with 0 below space leftwards harpoon over rightwards harpoon space straight M subscript straight s superscript plus space plus straight Y subscript straight s superscript minus Where comma space straight s equals space solubility space and space straight K subscript sp space equals solubility space product therefore space straight K subscript sp space equals straight s squared straight s equals square root of straight K subscript sp end root space equals space square root of 6.2 space straight x space 10 to the power of negative 13 end exponent end root space equals space 7.874 space straight x space 10 to the power of negative 7 end exponent simiarly comma space for space NY subscript 3 stack NY subscript 3 with 0 below space leftwards harpoon over rightwards harpoon stack space straight N subscript space superscript plus with straight s below space plus stack 3 straight Y to the power of minus with 3 straight s below space equals straight s space straight x space left parenthesis 3 straight s right parenthesis squared straight K subscript sp space equals 27 space straight s to the power of 4 straight s space equals space 4 square root of straight K subscript sp over 27 end root space equals space 4 square root of fraction numerator 6.2 space straight x space 10 to the power of negative 13 end exponent over denominator 27 end fraction end root space equals 3.89 space straight x space 10 to the power of negative 4 end exponent Therefore comma space molar space solubility space of space MY space in space water space is space less space than space that space of space NY subscript 3$

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18.

Match the compounds given in column I with the hybridization and shape given in column II and mark the correct option.

Column I	Column II
A.XeF₆	1.Distorted octahedral
B.XeO₃	2.Square planar
C.XeOF₄	3.Pyramidal
D.XeF₄	4.Square pyramidal

A

B

C

D

1 2 4 3
A

B

C

D

4 3 1 2
A

B

C

D

4 1 2 3
A

B

C

D

4 1 2 3

4 1 2 3

1051 Views

19.

Consider the following liquid-vapour equilibrium

$Liquid space rightwards harpoon over leftwards harpoon space Vapour$
Which of the following relations is correct?

$dlnP over dT equals fraction numerator negative increment H subscript v over denominator R T end fraction$
$dlnP over dT squared space equals space fraction numerator negative increment straight H subscript straight v over denominator straight T squared end fraction$
$dlnP over dT space equals space fraction numerator negative increment straight H subscript straight v over denominator RT squared end fraction$
$dlnP over dT space equals space fraction numerator negative increment straight H subscript straight v over denominator RT squared end fraction$

dlnP over dT space equals space fraction numerator negative increment straight H subscript straight v over denominator RT squared end fraction

The given phase equilibria is

$Liquid space rightwards harpoon over leftwards harpoon space Vapour$
This equilibrium states that, when liquid is heated, it converts into vapour but on cooling, it further converts into liquid, which is derived by Clausius clapeyron and the relationship is written as,

$dlnP over dT space equals space fraction numerator negative increment straight H subscript straight v over denominator RT squared end fraction where space increment straight H subscript straight v space equals space Heat space of space vaporisation$

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20.

When copper is heated with conc. HNO3 it produces

Cu(NO₃)₂ and NO
Cu(NO₃)₂, NO and NO₂
Cu(NO₃)₂ and N₂O
Cu(NO₃)₂ and N₂O

Cu(NO₃)₂ and N₂O

Nitric acid acts as an oxidising agent while reacting with copper.
i) when copper reacts with dilute nitric acid it forms,
3Cu + 4HNO₃(dilute) --> 3Cu(NO₃)₂ +2NO +4H₂O

ii) When copper reacts with concentrated nitric acid it forms,
Cu +4HNO₃(conc.) ---> Cu(NO₃)₂ +NO₂ +2H₂O