1
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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

Questions Asked from Some Basic Concepts of Chemistry

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