 JEE Mains Previous Years Questions with Solutions

4.5
star star star star star
1

JEE Main 2014 (Offline)

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)

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)

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

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

$\Rightarrow d = 1.22$