1
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.
2
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

3
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
4
MCQ (Single Correct Answer)

JEE Main 2017 (Online) 8th April Morning Slot

Excess of NaOH (aq) was added to 100 mL of FeCl3 (aq) resulting into 2.14 g of Fe(OH)3 . The molarity of FeCl3 (aq) is :

(Given molar mass of Fe = 56 g mol−1 and molar mass of Cl = 35.5 g mol−1)
A
0.2 M
B
03 M
C
0.6 M
D
1.8 M

Explanation

3 NaOH (aq.) + FeCl3(aq) $$ \to $$ Fe(OH)3(s)+ 3 NaCl(aq).

Moles of Fe(OH)3 = $${{2.14} \over {107}}$$ = 2 $$ \times $$ 10$$-$$2

1 mole of Fe(OH)3 is obtained from = 1 mole of FeCl3

$$\therefore\,\,\,$$ 2 $$ \times $$ 10$$-$$2 moles of Fe(OH)3 will obtain from

= 0.02 mole of FeCl3

Molarity of FeCl3 = $${{No.of\,moles} \over {Volume\,in\,L}}$$ = $${{2 \times {{10}^{ - 2}}} \over {0.1}}$$ = 0.2 M

Questions Asked from Some Basic Concepts of Chemistry

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