1

### JEE Main 2016 (Online) 9th April Morning Slot

5 L of an alkane requires 25 L of oxygen for its complete combustion. If all volumes are measured at constant temperature and pressure, the alkane is :
A
Ethane
B
Propane
C
Butane
D
Isobutane

## Explanation

We know, conbustion of Hydrocarbon

Cx Hy + (x + ${y \over 4}$)O2  $\to$  x CO2 + ${y \over 2}$ H2O

So, to consume 1 mole of Cx Hy we need

(x + ${y \over 4}$) mole of O2 gas.

As both Cx Hy and O2 are gas,

So avogadro's law is applicable on them.

At constant temperature and pressure according to avogadro's law, volume $\propto$ mole

So, for 5L of Cx Hy the

volume of O2 = 5(x + ${y \over 4}$)

According to the question,

5(x + ${y \over 4}$) = 25

$\therefore\,\,\,$ (x + ${y \over 4}$) = 5 . . . . . (1)

For Ethane (C2 H6), x = 2 and y = 6

$\therefore\,\,\,$ x + ${y \over 4}$ = 2 + ${6 \over 4}$ $\ne$ 5

$\therefore\,\,\,$ C2 H6 can't be the required alkane.

For Propane (C3 H8), x = 3. and y = 8

$\therefore\,\,\,$ x + ${y \over 4}$ = 3 + ${8 \over 4}$ = 5

$\therefore\,\,\,$ Propane (C3 H8) is the right alkane.

Similarly for Butane (C4 H10) and Isobutane (C4 H10) you can check
x + ${y \over 4}$ $\ne$ 5.
2

### JEE Main 2016 (Online) 10th April Morning Slot

The volume of 0.1N dibasic acid sufficient to neutralize 1 g of a base that furnishes 0.04 mole of OH in aqueous solution is :
A
200 mL
B
400 mL
C
600 mL
D
800 mL

## Explanation

According to law of equivalence,

Equivalence of acid = Equivalence of base.

Equivalence of acid = Normality x volume = 0.1 $\times$ v

As we know base produce OH$-$ ion, so moles of base is same as moles of OH$-$ ion = 0.04

Another formula of equivalence = n factor $\times$ number of moles

$\therefore\,\,\,$ Equivalance of base = n factor of OH$-$ $\times$ moles of OH$-$ = 1 $\times$ 0.04

As for any ion, the charge of that ion is the n factor of that ion. Here OH$-$ has 1 negative charge so it's n factor = 1

$\therefore\,\,\,$ 0.1 $\times$ v = 1 $\times$ 0.04

$\Rightarrow$$\,\,\,$ v = 0.4 L

=   0.4 $\times$ 1000

=    400 ml.
3

### JEE Main 2017 (Offline)

1 gram of a carbonate (M2CO3) on treatment with excess HCl produces 0.01186 mole of CO2. The molar mass of M2CO3 in g mol–1 is:
A
84.3
B
118.6
C
11.86
D
1186

## Explanation

M2CO3 + 2HCl $\buildrel \, \over \longrightarrow$ 2MCl + CO2 + H2O

Here weight of M2CO3 = 1 gm

Let molar mass of M2CO3 = M

$\therefore\,\,\,$ No of moles of M2CO3 = ${1 \over M}$

For this reaction,

${{{1 \over M}} \over 1} = {{0.01186} \over 1}$

$\Rightarrow \,\,\,M$ = ${1 \over {0.01186}}$ = 84.3 g mol-1

NOTE :

For any reaction this will be always valid.

For this reaction, 2KClO3 $\buildrel \, \over \longrightarrow$ 2KCl + 3O2

We can write, ${{{n_{KCl{O_3}}}} \over 2} = {{{n_{KCl}}} \over 2} = {{{n_{{O_2}}}} \over 3}$

This means 4

### JEE Main 2017 (Offline)

The most abundant elements by mass in the body of a healthy human adult are: Oxygen (61.4%); Carbon (22.9%), Hydrogen (10.0%); and Nitrogen (2.6%). The weight which a 75 kg person would gain if all 1H atoms are replaced by 2H atoms is:
A
37.5 kg
B
7.5 kg
C
10 kg
D
15 kg

## Explanation

Given that weight of human adult is = 75 kg

Among those 75 kg, 10% is Hydrogen(1H).

$\therefore$ Mass of 1H = $75 \times {{10} \over {100}}$ = 7.5 kg

Now when every 1H atom is replaced by 2H atom then weight of every atom is become double. So total weight of 2H becomes = 2$\times$7.5 = 15 kg.

So the weight gain by the person = 15 - 7.5 = 7.5 kg