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JEE Mains Previous Years Questions with Solutions

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1

JEE Main 2014 (Offline)

MCQ (Single Correct Answer)
The ratio of masses of oxygen and nitrogen in a particular gaseous mixture is 1 : 4. The ratio of number of their molecule is:
A
1: 8
B
3 : 16
C
1 : 4
D
7 : 32

Explanation

Given ratio of mass of O2 and N2 = 1 : 4
Let mass of O2 = w
and mass of N2 = 4w

$$\therefore$$ Number of moles of O2 = $${w \over {32}}$$
Number of molecules of O2 = $${w \over {32}}$$$$ \times $$NA

Number of moles of N2 = $${4w \over {28}}$$
Number of molecules of N2 = $${4w \over {28}}$$$$ \times $$NA

$$\therefore$$ Ratio of molecules of O2 and N2
= $${w \over {32}}$$$$ \times $$NA : $${4w \over {28}}$$$$ \times $$NA

= 7 : 32
2

JEE Main 2013 (Offline)

MCQ (Single Correct Answer)
The molarity of a solution obtained by mixing 750 mL of 0.5 (M) HCl with 250 mL of 2(M) HCl will be:
A
1.00 M
B
1. 75 M
C
0.975 M
D
0.875 M

Explanation

The formula for molarity of mixture of two substance is = $${{{M_1}{V_1} + {M_2}{V_2}} \over {{V_1} + {V_2}}}$$

Here $${M_1} = 0.5$$, $${V_1} = 750$$, $${M_2} = 2$$, $${V_2} = 250$$

$$\therefore$$ Molarity of mixture = $${{0.5 \times 750 + 2 \times 250} \over {750 + 250}}$$

= $${{375 + 500} \over {1000}}$$

= $${{875} \over {1000}}$$

= 0.875
3

JEE Main 2013 (Offline)

MCQ (Single Correct Answer)
A gaseous hydrocarbon gives upon combustion 0.72 g of water and 3.08 g of CO2. The empirical formula of the hydrocarbon is
A
C2H4
B
C3H4
C
C6H5
D
C7H8

Explanation

Required reaction,

C$$x$$H$$y$$ + $$\left( {x + {y \over 4}} \right)$$O2 $$ \to $$$$x$$CO2 + $${{y \over 2}}$$H2O

0.72 gm of H2O = $${{{0.72} \over {18}}}$$ mole of H2O = 0.04 mole of H2O

In one H2O molecule 2 hydrogen atoms present.

So in 0.04 mole of H2O molecules 2$$ \times $$0.04 = 0.08 moles of H atoms present.

0.72 gm of CO2 = $${{{3.08} \over {44}}}$$ mole of CO2 = 0.07 mole of CO2

And in one CO2 molecule 1 C atom present.

So in 0.07 mole of CO2 molecules 0.07$$\times$$1 = 0.07 moles of C atoms present

$$\therefore$$ C : H = 0.07 : 0.08 = 7 : 8

$$\therefore$$ Empirical formula of hydrocarbon = C7H8
4

AIEEE 2007

MCQ (Single Correct Answer)
The density (in g mL-1) of a 3.60 M sulphuric acid solution that is 29% H2SO4 (molar mass = 98 g mol-1) by mass will be
A
1.45
B
1.64
C
1.88
D
1.22

Explanation

Important formula of Molarity(M) when % w/w is given

M = $${{10 \times \% w/w \times d} \over {{M_{solute}}}}$$

Here M = 3.6, % w/w = 29, d = density, Msolute = 98

$$\therefore$$ 3.6 = $${{10 \times 29 \times d} \over {98}}$$

$$ \Rightarrow d = 1.22$$

Questions Asked from Some Basic Concepts of Chemistry

On those following papers in MCQ (Single Correct Answer)
Number in Brackets after Paper Indicates No. of Questions
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JEE Main 2019 (Online) 12th April Evening Slot (2)
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JEE Main 2019 (Online) 12th January Evening Slot (1)
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JEE Main 2019 (Online) 11th January Evening Slot (1)
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JEE Main 2019 (Online) 11th January Morning Slot (1)
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JEE Main 2019 (Online) 10th January Evening Slot (1)
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JEE Main 2019 (Online) 9th January Evening Slot (1)
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JEE Main 2019 (Online) 9th January Morning Slot (2)
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JEE Main 2018 (Online) 16th April Morning Slot (1)
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JEE Main 2018 (Offline) (1)
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JEE Main 2018 (Online) 15th April Morning Slot (1)
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JEE Main 2017 (Online) 9th April Morning Slot (1)
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JEE Main 2017 (Online) 8th April Morning Slot (1)
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JEE Main 2017 (Offline) (2)
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JEE Main 2016 (Online) 10th April Morning Slot (1)
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JEE Main 2016 (Online) 9th April Morning Slot (3)
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JEE Main 2014 (Offline) (1)
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JEE Main 2013 (Offline) (2)
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AIEEE 2007 (2)
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AIEEE 2006 (2)
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AIEEE 2005 (2)
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AIEEE 2004 (3)
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AIEEE 2003 (2)
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AIEEE 2002 (3)
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