Some Basic Concepts Of Chemistry - Mole Concept — DPP 1

20 NEET questions from Some Basic Concepts in Chemistry, with answers and explanations. Attempt it as a timed test to get scoring, accuracy breakdown and mistake tracking.

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1
The weight of one atom of Uranium is 238 amu. Its actual weight is .... g.
  1. A $1.43 \times 10 ^ { 26 }$
  2. B $3.94 \times 10 ^ { - 22 }$
  3. C $6.99 \times 10 ^ { - 23 }$
  4. D $1.53 \times 10 ^ { - 22 }$
Show answer & explanation

Correct answer: B

1 “U” atom = 238 amu
1 amu = 1.66 × 10$^{–24}$g
1 “U” atom = 238 × (1.66 × 10$^{-24}$g)
≃ 3.94 × 10$^{–22}$g
2
Which of the following has the highest mass?
  1. A 1 g atom of C
  2. B $1 / 2$ mole of $\mathrm { CH } _ { 4 }$
  3. C 10 mL of water
  4. D $3.011 \times 10 ^ { 23 }$ atoms of oxygen
Show answer & explanation

Correct answer: A

1g atom of C = 1 mole of C
= 12g
3
Which of the following contains maximum number of oxygen atoms?
  1. A 1 g of O
  2. B 1 g of $\mathrm { O } _ { 2 }$
  3. C 1 g of $\mathrm { O } _ { 3 }$
  4. D all have the same number of atoms
Show answer & explanation

Correct answer: D

Some Basic Concepts Of Chemistry - Mole Concept – NEET-5633 All have the same number of atoms.
4
A gas is found to have the formula (CO) ${ } _ { \mathrm { x } }$. It's VD is 70 . The value of $x$ must be:-
  1. A 7
  2. B 4
  3. C 5
  4. D 6
Show answer & explanation

Correct answer: C

Formula of gas = [CO]$_{ x}$
Mol. wt. of gas = [12 + 16]$_{ x}$
= 28x
28x = 2 × V.D.
28x = 2 × 70
28x = 140
x = $\frac{140}{28}$ = 5
5
Number of neutrons present in 1.7 g of ammonia is -
  1. A $\mathrm { N } _ { \mathrm { A } }$
  2. B $\mathrm { N } _ { \mathrm { A } } / 10 \times 4$
  3. C $\left( \mathrm { N } _ { \mathrm { A } } / 10 \right) \times 7$
  4. D $\mathrm { N } _ { \mathrm { A } } \times 10 \times 7$
Show answer & explanation

Correct answer: C

$\begin{array}{l}\text { Number of neutrons } \\ =\mathrm{n} \times \mathrm{N}_{\mathrm{A}} \times \text { (number of neutrons) } \\ =\frac{1.7}{17} \times \mathrm{N}_{\mathrm{A}} \times 7 \\\end{array}$
$=\frac{\mathrm{N}_{\mathrm{A}}}{10} \times 7$
6
Vapour density of gas is 11.2. Volume occupied by 2.4 g of this at STP will be -
  1. A 11.2 L
  2. B 2.24 L
  3. C 22.4 L
  4. D 2.4 L
Show answer & explanation

Correct answer: D

Mol. wt. = 2 × 11.2 = 22.4g
2.4 moles of gas = $\frac{2.4}{22.4}$
Vol. of gas (at STP) = moles × 22.42
= $\frac{2.4}{22.4}$
= 22.4
7
Which of the following contains maximum number of atoms?
  1. A 4 g of $\mathrm { H } _ { 2 }$
  2. B 16 g of $\mathrm { O } _ { 2 }$
  3. C $28 \mathrm {~g} \mathrm { of } \mathrm { N } _ { 2 }$
  4. D 18 g of $\mathrm { H } _ { 2 } \mathrm { O }$
Show answer & explanation

Correct answer: A

Some Basic Concepts Of Chemistry - Mole Concept – NEET-5635
8
The actual weight of a molecule of water is -
  1. A 18 g
  2. B $2.99 \times 10 ^ { - 23 } \mathrm {~g}$
  3. C both (A) & (B) are correct
  4. D 1.66 × 10$^{–24}$ g
Show answer & explanation

Correct answer: B

1 H₂O molecule = 18 amu
Actual mass of 1 H₂O molecule
= 18 × (1.66 × 10$^{–24}$g)
≃ 2.99 × 10$^{–23}$
9
What is the mass of a molecule of $\mathrm { CH } _ { 4 }$ :-
  1. A 16 g
  2. B $26.6 \times 10 ^ { 22 } \mathrm {~g}$
  3. C $2.66 \times 10 ^ { - 23 } \mathrm {~g}$
  4. D $16 \mathrm {~N} _ { \mathrm { A } } $ g
Show answer & explanation

Correct answer: C

1 molecule CH₄ = 16 amu
1 amu = (1.66 × 10$^{–24}$g)
mass of 1 molecule of CH₄ = 16 × (1.66 × 10$^{–24}$g)
= 2.66 × 10$^{–23}$g
10
If VmL of the vapours of substance at NTP weight W g. Then molecular weight of substance is:-
  1. A ( W / V ) $\times$ 22400
  2. B $\frac { \mathrm { V } } { \mathrm { W } } \times 22.4$
  3. C ( W - V ) $\times$ 22400
  4. D $\frac { \mathrm { W } \times 1 } { \mathrm {~V} \times 22400 }$
Show answer & explanation

Correct answer: A

mass of gas = w$_{(g)}$
Vol. of gas = V mL (at STP)
mass w moles of gas = $\frac{\text { mass }}{\text { mol.wt. }}=\frac{\mathrm{w}}{\text { mol.wt. }}$

moles of gas =$\frac{\mathrm{V}_{\mathrm{STP}}(\mathrm{ml})}{22400}$

$\frac{\mathrm{w}}{\text { mol.wt. }}=\frac{\mathrm{V}_{\text {STP }}(\mathrm{ml})}{22400}$

Mol. wt. = $\frac{\mathrm{w}}{\mathrm{V}_{(\mathrm{ml})}} \times 22400$
11
The number of electron in $3.1 \mathrm {~mg }$ $\mathrm { NO } _ { 3 } { } ^ { - }$is ( $\mathrm { N } _ { \mathrm { A } } = 6 \times 10 ^ { 23 }$ )
  1. A 32
  2. B $1.6 \times 10 ^ { - 3 }$
  3. C $9.6 \times 10 ^ { 20 }$
  4. D $9.6 \times 10 ^ { 23 }$
Show answer & explanation

Correct answer: C

Number of electrons in
$\mathrm{NO}_{3}^{-}=\mathrm{n}_{\mathrm{NO}_{3}^{-}} \times \mathrm{N}_{\mathrm{A}}$ (number of electrons)
$=\frac{3.1 \times 10^{-3}}{62} \times 6.022 \times 10^{23} \times 32$
= 9.6 × 10$^{20}$
12
4.4 g of an unknown gas occupies 2.24 L of volume at STP. The gas may be :-
  1. A $\mathrm { N } _ { 2 } \mathrm { O }$
  2. B CO
  3. C $\mathrm { CO } _ { 2 }$
  4. D (A) & (C) both
Show answer & explanation

Correct answer: D

n$_{gas}$ = $\frac{2.24}{22.4}$
Molar mass = $\frac{\mathrm{w}_{\mathrm{gas}}}{\mathrm{n}_{\mathrm{gas}}}$
= $\frac{4.4}{0.1}$ = 44
CO₂ & N₂O
13
The volume of a gas in discharge tube is $1.12 \times 10 ^ { - 7 } \mathrm {~mL}$ at STP. Then the number of molecule of gas in the tube is -
  1. A $3.01 \times 10 ^ { 4 }$
  2. B $3.01 \times 10 ^ { 15 }$
  3. C $3.01 \times 10 ^ { 12 }$
  4. D $3.01 \times 10 ^ { 16 }$
Show answer & explanation

Correct answer: C

$\begin{aligned} n_{\mathrm{gas}} & =\frac{V_{\text {STP }}(\mathrm{ml})}{22400} \\ & =\frac{1.12 \times 10^{-7}}{22400}=\frac{1}{2} \times 10^{-11}\end{aligned}$
molecules of gas = moles × N$_{A}$
= $\frac{1}{2}$ × 10$^{-11}$ × 6.02 × 10$^{23}$
= 3.01 × 10$^{12}$
14
Elements $\mathrm { A }$ and $\mathrm { B }$ form two compounds $\mathrm { B } _ { 2 } \mathrm {A} _ { 3 }$ and $\mathrm { B } _ { 2 } \mathrm {A}$. 0.05 moles of $\mathrm { B } _ { 2 } \mathrm {A} _ { 3 }$ weight 9.0 g and 0.10 mole of $\mathrm { B } _ { 2 } \mathrm {A}$ weight 10 g . Calculate the atomic weight of A and B :-
  1. A 20 and 30
  2. B 30 and 40
  3. C 40 and 30
  4. D 30 and 20
Show answer & explanation

Correct answer: C

Let atomic wt. of A = a
Atomic wt. of B = b

Molar mass of
B$_{2}$A$_{3}$ = (2 × b) + (3 × a)
= mass of 1 mole substance
0.05 mol B$_{2}$A$_{3}$ → 9g
1 mol of B$_{2}$A$_{3}$ = $\frac{9}{0.05}$ =180g
2b + 3a = 180 ……(1)

molar mass of B$_{2}$A = (2b + a)
0.1 mol of B$_{2}$A → 10g
1 mol of B$_{2}$A → $\frac{10}{0.1}$ = 100
⇒ 2b + a = 100 ……(2)
On solving a = 40, b = 30
15
Number of oxygen atoms in 8 g of ozone is -
  1. A $6.02 \times 10 ^ { 23 }$
  2. B $\frac { 6.02 \times 10 ^ { 23 } } { 2 }$
  3. C $\frac { 6.02 \times 10 ^ { 23 } } { 3 }$
  4. D $\frac { 6.02 \times 10 ^ { 23 } } { 6 }$
Show answer & explanation

Correct answer: B

Number of atoms in $\left[\mathrm{O}_3\right]$
= $\mathrm{n} \times \mathrm{N}_{\mathrm{A}} \times$ number of atoms
$=\frac{8}{48} \times \mathrm{N}_{\mathrm{A}} \times 3$
$=\frac{\mathrm{N}_{\mathrm{A}}}{2}=\frac{6.022 \times 10^{23}}{2}$
16
The number of atoms present in 0.5 g atom of nitrogen is same as the atoms in -
  1. A 12 g of C
  2. B 32 g of S
  3. C 8 g of oxygen
  4. D 24 g of Mg
Show answer & explanation

Correct answer: C

Number of atom in 0.5 g atom of nitrogen
$=0.5 \times \mathrm{N_{A}} \times 2=\mathrm{N_{A}}$
Number of atom in 8 g oxygen
$=\frac{8}{16} \times \mathrm{N_{A}} \times 2=\mathrm{N_{A}}$
17
The number of atoms present in 16 g of oxygen is
  1. A $6.02 \times 10 ^ { 11.5 }$
  2. B $3.01 \times 10 ^ { 23 }$
  3. C $3.01 \times 10 ^ { 11.5 }$
  4. D $6.02 \times 10 ^ { 23 }$
Show answer & explanation

Correct answer: D

$\begin{array}{l}16 \mathrm{~g} \text { oxygen }\left(\mathrm{O}_{2}\right) \\ \mathrm{n}_{\mathrm{O}_{2}}=\frac{16}{32}=\frac{1}{2} \\ \text { Number of O-atoms } \\ =\mathrm{n}_{\mathrm{O}_{2}} \times \mathrm{N}_{\mathrm{A}} \times \text { atomicity } \\ =\frac{1}{2} \times 6.022 \times 10^{23} \times 2 \\ =6.022 \times 10^{23}\end{array}$
18
Which of the following contains the least number of molecules?
  1. A $4.4 \mathrm {~g}$ $\mathrm { CO } _ { 2 }$
  2. B $3.4 \mathrm {~g}$ $\mathrm { NH } _ { 3 }$
  3. C $1.6 \mathrm {~g}$ $\mathrm { CH } _ { 4 }$
  4. D $3.2 \mathrm {~g}$ $\mathrm { SO } _ { 2 }$
Show answer & explanation

Correct answer: D

Some Basic Concepts Of Chemistry - Mole Concept – NEET-5644
19
Sum of number of protons, electrons and neutrons in 12 g of ${ } _ { 6 } ^ { 12 } \mathrm { C }$ is :-
  1. A 1.8
  2. B $12.044 \times 10 ^ { 23 }$
  3. C $1.084 \times 10 ^ { 25 }$
  4. D $10.84 \times 10 ^ { 23 }$
Show answer & explanation

Correct answer: C

Sum of protons, electrons and neutrons in 12 g of ${ }_{6}^{12} \mathrm{C}$
= n $\times \mathrm{N_{A}} \times$ [sum of e, p & n]
$=\frac{12}{12} \times \mathrm{N_{A}} \times 18$
= 18 × 6.022 × 10²³
= 1.08 × 10²⁵
20
5.6 L of oxygen at STP contains -
  1. A $6.02 \times 10 ^ { 23 }$ atoms
  2. B $3.01 \times 10 ^ { 23 }$ atoms
  3. C $1.505 \times 10 ^ { 23 }$ atoms
  4. D $0.7525 \times 10 ^ { 23 }$ atoms
Show answer & explanation

Correct answer: B

$\begin{array}{l}\text { Number of atoms of oxygen } \\ =n \times \mathrm{N_{A}} \times \text { number of atoms } \\ =\frac{5.6}{22.4} \times 6.022 \times 10^{23} \times 2 \\ =3.01 \times 10^{23} \text { atoms. }\end{array}$

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