In this article, JEE aspirants will get Solved Chemistry Practice Paper based on the latest pattern and covering most of the syllabus of Chemistry for JEE Main 2018. It will help students to track their progress and perform well in the examination.
Benefits of solving this practice paper:
1. Students will get familiar with the pattern and difficulty level of the exam.
2. Students can track their preparation level of JEE Main 2018.
3. Students can practise effective revision strategies.
4. JEE aspirants will learn new strategies to manage time and accuracy in the examination.
6. Students can improve their knowledge by practicing new set of questions from the complete syllabus of Chemistry.
JEE Main Chemistry Syllabus 2018
About the paper:
1. This paper contains 30 objective type questions with only one correct option.
2. The questions have been taken from different topics like Basic Concepts of Chemistry, Structure of Atom, Classification of Elements & Periodicity in Properties, Chemical Bonding and Molecular Structure, Thermodynamics, Equilibrium, Redox Reactions, Hydrocarbons, Solid State, Solutions, Electrochemistry, Chemical Kinetics, Surface Chemistry, Isolation of Elements, Coordination Compounds, Haloalkanes and Haloarenes, Alcohols, Phenols and Ethers, Aldehydes, Ketones and Carboxylic Acids, Organic Compounds containing Nitrogen, Biomolecules, Polymers.
3. Detail solution for all the questions.
Few sample questions from the Practice Paper are given below:
Question:
Which of the following represents the correct order of increasing first ionization enthalpy for Ca, Ba, S, Se and Ar?
(a) Ca < S < Ba < Se < Ar
(b) S < Se < Ca < Ba < Ar
(c) Ba < Ca < Se < S < Ar
(d) Ca < Ba < S < Se < Ar
Sol. (c)
It is known that ionisation energy increase along a period from left to right and decreases down a group.
The position of given elements in the periodic table is as follows:
First row represent the group and remaining cells represent elements.
| 2 |
16 |
8 |
| Ca |
S |
Ar |
| Ba |
Se |
|
So, correct order of increasing first ionisation enthalpy is as shown below:
Ba < Ca < Se < S < Ar
Question:
In SN2 reactions, the correct order of reactivity for the following compounds CH3Cl, CH3CH2Cl, (CH3)2 CHCl and (CH3)3CCl is
(a) CH3Cl > (CH3)2CHCl > CH3CH2Cl > (CH3) 3CCl
(b) CH3CI > CH3CH2 CI > (CH3)2 CHCl > (CH3)3CCl
(c) CH3CH2Cl > CH3Cl > (CH3)2CHCl > (CH3) 3CCl
(d) (CH3)2CHCI > CH3CH2Cl > CH3Cl > (CH3)3CCl
Sol. (b)
In SN2 reactions, the order of reactivity is followed as:
Primary halide > Secondary halide > Tertiary halide
So, the order of reactivity of the given compound is followed as:
CH3CI > CH3CH2 CI > (CH3)2 CHCl > (CH3)3CCl
Question:
For the process, H2O(l) H2O (g) at T = 100° C and 1 atmosphere pressure, the correct choice is
(a) ΔS system > 0 and ΔS surrounding > 0
(b) ΔS system > 0 and ΔS surrounding < 0
(c) ΔS system < 0 and ΔS surrounding > 0
(d) ΔS system < 0and ΔS surrounding < 0
Sol. (d)
We know that total entropy change of universe is zero.
Given,
At 100°C and 1 atmosphere pressure,
H2O (l) H2O (g) is at equilibrium.
For equilibrium, ΔS total = 0
and ΔS system + ΔS surrounding = 0
During conversion of liquid to gas the entropy of system increases and in similar manner entropy of surrounding decreases.
∴ ΔS system > 0 and ΔS surrounding < 0
Question:
Metallic radii of some transition elements are given below. Which of these elements will have highest density?
| Element |
Fe |
Co |
Ni |
Cu |
| Metallic Radii/pm |
126 |
125 |
125 |
128 |
(a) Fe
(b) Ni
(c) Co
(d) Cu
Sol. (d)
Density is measured as mass per unit volume. On moving from left to right in a period of periodic table, metallic radius decreases while the atomic mass increases. Decrease in metallic radius means decreased volume. Thus decrease in volume coupled with the increase in atomic mass results in increase in density of metal. Hence, the density of elements increases on moving from left to right in a period. Thus Cu among all the given metals, being present at the extreme right of the period, is having the highest density.
Question:
Which of the following compounds is detected by Molisch’s test?
(a) Sugars
(b) Amines
(c) Primary alcohols
(d) Nitro compounds
Sol. (a)
Molisch's test is a confirmatory test for carbohydrates or sugar. A few drops of alcoholic solution of are added to a carbohydrate solution. Then 1 ml of concentrated sulphuric acid is added from the sides of the test tube. Appearance of violet ring at the junction of two liquids confirms the presence of carbohydrate or sugar.
Question:
On addition of 1 mL solution of 10% NaCl to 10mL of gold sol in the presence of 0.25 g of starch, the coagulation is just prevented. The gold number of starch is:
(a) 0.25
(b) 2.5
(c) 25.0
(d) 250
Sol. (d)
Gold number of a protective colloid is the minimum mass of it in milligrams which must be added to 10 mL of gold sol so that no coagulation of the gold sol takes place when 1 mL of 10% NaCl solution is added to it.
Now given, starch added to 10 mL of gold soil to completely prevent its coagulation by 10 mL of 10% NaCl solution = 0.25 g = 250 mg
Conclusion:
This practice paper will help students to practice new questions from the complete syllabus of Chemistry. Students can easily track their progress after solving this practice paper. Also they can revise the topics where they stuck while attempting this paper.
Download Complete Practice Paper
Also Read
