Some Basic Concepts of Chemistry JEE JEE Main| JEE Main Important Questions

Q 1 :

0.05 cm thick coating of silver is deposited on a plate of 0.05 $m^{2}$ area. The number of silver atoms deposited on plate are _________ $\times 10^{23}$ . (At mass Ag = 108, d = 7.9 g $c m^{- 3}$ ) [2024]

(11) Volume of silver = thickness $\times$ area of plate

= 0.05 cm $\times$ 0.05 $m^{2}$

= 0.05 cm $\times$ 0.05 $\times$ ${(100 c m)}^{2}$

= 25 $c m^{3}$

Mass of silver deposited = density $\times$ volume

= 7.9 $g c m^{- 3} \times$ 25 $c m^{3}$

= 197.5 g

Moles of silver deposited(n) = $\frac{M a s s o f s i l v e r d e p o s i t e d}{m o l a r m a s s o f s i l v e r}$

= $\frac{197.5 g}{108 g m o l^{- 1}}$

= 1.8287 mol

Atoms of silver deposited = $n \times N_{A}$ atoms

= 1.8287 $\times$ 6.022 $\times 10^{23}$

= 11.01 $\times 10^{23}$ atoms

Q 2 :

$2.8 \times 10^{- 3}$ mol of ${CO}_{2}$ is left after removing $10^{21}$ molecules from its $‘ x ’$ mg sample. The mass of ${CO}_{2}$ taken initially is

Given: $N_{A} = 6.02 \times 10^{23} {mol}^{- 1}$ [2025]

48.2 mg
98.3 mg
150.4 mg
196.2 mg

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(4)

$10^{21} molecules = \frac{10^{21}}{6.02 \times 10^{23}} mol = 1.66 \times 10^{- 3} mol$

Initial moles = mole left + mole removed

Initial moles $= 2.8 \times 10^{- 3} mol + 1.66 \times 10^{- 3} mol$

$= 4.46 \times 10^{- 3} mol$

$Initial mass = Initial moles \times molar mass$

$= 4.46 \times 10^{- 3} mol \times 44 {g mol}^{- 1}$

$= 196.24 \times 10^{- 3} g = 196.24 mg$

Q 3 :

Among $10^{- 9}$ g (each) of the following elements, which one will have the highest number of atoms?

Element: Pb, Po, Pr and Pt [2025]

Po
Pr
Pb
Pt

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(2)

$Number of atoms = \frac{Given mass}{Atomic mass}$

As given mass is same for the four elements,

$number of atoms \propto \frac{1}{Atomic mass} .$

Po and Pb are present in 6th period, p block. Pt is present in 5d series. Pr is present in 4f series. As Pr is placed before Po, Pb and Pt in periodic table, it has least atomic mass. As it has least atomic mass, it has most number of atoms.

Q 4 :

The number of molecules and moles in $2.8375 litres$ of $O_{2}$ at STP are respectively [2023]

$7.527 \times 10^{22}$ and $0.125 mol$
$1.505 \times 10^{23}$ and $0.250 mol$
$7.527 \times 10^{23}$ and $0.125 mol$
$7.527 \times 10^{22}$ and $0.250 mol$

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(1)

$No. of moles = \frac{Volume at S.T.P (L)}{22.4} = \frac{2.8375}{22.4} = 0.125 mole$

$No. of molecules = no. of moles \times N_{A}$

$= 0.125 \times 6.023 \times 10^{23}$

$= 7.527 \times 10^{22}$

Q 5 :

Match List I with List II. [2023]

List I List II

A. 16 g of ${CH}_{4} (g)$ I. Weighs 28 g

B. 1 g of $H_{2} (g)$ II. $60.2 \times 10^{23}$ electrons

C. 1 mole of $N_{2} (g)$ III. Weighs 32 g

D. 0.5 mol of ${SO}_{2} (g)$ IV. Occupies 11.4 L volume at STP

Choose the correct answer from the options given below:

A-II, B-IV, C-III, D-I
A-I, B-III, C-II, D-IV
A-II, B-III, C-IV, D-I
A-II, B-IV, C-I, D-III

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(4)

(A) $16 g or 1 mol C H_{4} or 6.02 \times 10^{23} \times 10 electrons$

$= 60.2 \times 10^{23} electrons (II)$

(B) ${1 g H}_{2} or \frac{1}{2} mol H_{2} or 11.35 L volume at STP (IV)$

(C) $1 mol N_{2} = 28 g N_{2} (I)$

(D) $0.5 mol S O_{2} = 32 g S O_{2} (III)$

Q 6 :

In the reaction,

$2 Al (s) + 6 HCl (aq) \to 2 {Al}^{3 +} (aq) + 6 {Cl}^{-} (aq) + 3 H_{2} (g)$ [2026]

$33.6 L H_{2} (g)$ is produced regardless of temperature and pressure for every mole of Al that reacts.
$12 L HCl (aq)$ is consumed for every $6 L H_{2} (g)$ produced.
$67.2 L H_{2} (g)$ at STP is produced for every mole of Al that reacts.
$11.2 L H_{2} (g)$ at STP is produced for every mole of HCl consumed.

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(4)

$2 Al (s) + 6 HCl (aq) \to 2 {Al}^{3 +} (aq) + 6 {Cl}^{-} (aq) + 3 H_{2} (g)$

$Mole of H_{2} produced$

$= 2 \times mole of HCl used$

$= \frac{2}{3} \times mole of Al used$

Q 7 :

Number of moles and number of molecules in 1.4187 L of $S O_{2}$ at STP respectively are [2026]

$0.1266; 3.812 \times 10^{22}$
$0.0633; 3.812 \times 10^{22}$
$0.1266; 7.6238 \times 10^{22}$
$0.0633; 7.6238 \times 10^{22}$

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(2)

Number of moles $(n)$

$= \frac{Volume at STP in L}{22.4} = \frac{1.4187}{22.4} mol = 0.0633 mol$

Number of molecules (N)

$= n \times N_{A} = 0.0633 \times 6.023 \times 10^{23} = 3.812 \times 10^{22}$

Q 8 :

The correct order of total number of atoms present in

(A) 2 moles of cyclohexane

(B) 684 g of sucrose

(C) 90.8 L of dihydrogen at STP

is: [2026]

C > A > B
C > B > A
B > C > A
B > A > C

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(4)

$(A) 2 mol cyclohexane$ $(C_{6} H_{12})$ $molecules$

$=$ $2 \times (6 + 12)$ $mol atoms = 36 mol atoms$

$(B) 684 g sucrose (C_{12} H_{22} O_{11}) = (\frac{684}{342}) mol sucrose molecules$

$= (\frac{684}{342}) \times (12 + 22 + 11) mol atoms = 90 mol atoms$

$(C) H_{2} molecules in 90.8 {L H}_{2} at STP$

$= \frac{90.8}{22.7} mol H_{2} molecules$

$= \frac{90.8}{22.7} \times 2 mol H atoms = 8 mol H atoms.$

$Correct order of number of atoms = B > A > C$

Q 9 :

Which of the following contain the same number of atoms?

(Given: Molar mass in g ${mol}^{- 1}$ of H, He, O and S are 1, 4, 16 and 32 respectively)

A. 2 g of $O_{2}$ gas

B. 4 g of ${SO}_{2}$ gas

C. 1400 mL of $O_{2}$ at STP

D. 0.05 L of He at STP

E. 0.0625 mol of $H_{2}$ gas

Choose the correct answer from the options given below: [2026]

A and B only
B and C only
C and D only
A, C and E only

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(4)

Substance	Moles of substance (given mass/molar mass)	Number of molecules	Atoms = Atomicity × number of molecules
(A) 2 g of $O_{2}$ gas	$\frac{2}{32} = \frac{1}{16}$	$\frac{1}{16} N_{A}$	$2 \times \frac{1}{16} N_{A} = \frac{1}{8} N_{A} = 0.125 N_{A}$
(B) 4 g of ${SO}_{2}$ gas	$\frac{4}{64} = \frac{1}{16}$	$\frac{1}{16} N_{A}$	$3 \times \frac{1}{16} N_{A} = 0.1875 N_{A}$
(E) $H_{2}$ gas	$0.0625$	$0.0625 N_{A}$	$2 \times 0.0625 N_{A} = 0.125 N_{A}$

Volume of gas at STP	Moles of substance (Vol in L / 22.4)	Number of molecules	Atoms = Atomicity × number of molecules
(C) 1400 mL (1.4 L) of $O_{2}$ gas	$\frac{1.4}{22.4} = \frac{1}{16}$	$\frac{1}{16} N_{A}$	$2 \times \frac{1}{16} N_{A} = \frac{1}{8} N_{A} = 0.125 N_{A}$
(E) 0.05 L of He gas	$\frac{0.05}{22.4} = \frac{1}{448}$	$2 \times \frac{1}{16} N_{A} = \frac{1}{8} N_{A}$	$\frac{1}{448} N_{A} = 0.00223 N_{A}$

Q 10 :

Match List – I with List – II.

List – I List – II

(Mass of substance) (Number of atoms)

A. 1.8 mg water I. $2 \times 10^{- 4} \times N_{A}$

B. 9.8 mg sulphuric acid II. $1.5 \times 10^{- 4} \times N_{A}$

C. 1.8 mg carbon III. $3 \times 10^{- 4} \times N_{A}$

D. 5.85 mg salt (NaCl) IV. $7 \times 10^{- 4} \times N_{A}$

Choose the correct answer from the options given below: [2026]

A-IV, B-III, C-I, D-II
A-III, B-II, C-IV, D-I
A-III, B-IV, C-II, D-I
A-III, B-IV, C-I, D-II

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(3)

Mass of substance	Moles of substance (given mass/molar mass)	Number of molecules	Atoms = Atomicity × number of molecules
1.8 mg water $(H_{2} O)$	$\frac{1.8 \times 10^{- 3}}{18} = 10^{- 4}$	$10^{- 4} N_{A}$	$3 \times 10^{- 4} N_{A}$
9.8 mg sulphuric acid $(H_{2} {SO}_{4})$	$\frac{9.8 \times 10^{- 3}}{98} = 10^{- 4}$	$10^{- 4} N_{A}$	$7 \times 10^{- 4} N_{A}$
1.8 mg carbon (C)	$\frac{1.8 \times 10^{- 3}}{12} = 1.5 \times 10^{- 4}$	$1.5 \times 10^{- 4} N_{A}$	$1.5 \times 10^{- 4} N_{A}$
5.85 mg NaCl	$\frac{5.85 \times 10^{- 3}}{58.5} = 10^{- 4}$	$10^{- 4} N_{A}$	$2 \times 10^{- 4} N_{A}$

Topic Question Set

Q 1 :

0.05 cm thick coating of silver is deposited on a plate of 0.05 $m^{2}$ area. The number of silver atoms deposited on plate are _________ $\times 10^{23}$ . (At mass Ag = 108, d = 7.9 g $c m^{- 3}$ ) [2024]

Q 2 :

$2.8 \times 10^{- 3}$ mol of ${CO}_{2}$ is left after removing $10^{21}$ molecules from its $‘ x ’$ mg sample. The mass of ${CO}_{2}$ taken initially is

Given: $N_{A} = 6.02 \times 10^{23} {mol}^{- 1}$ [2025]

Q 3 :

Among $10^{- 9}$ g (each) of the following elements, which one will have the highest number of atoms?

Element: Pb, Po, Pr and Pt [2025]

Q 4 :

The number of molecules and moles in $2.8375 litres$ of $O_{2}$ at STP are respectively [2023]

Q 6 :

In the reaction,

$2 Al (s) + 6 HCl (aq) \to 2 {Al}^{3 +} (aq) + 6 {Cl}^{-} (aq) + 3 H_{2} (g)$ [2026]

Q 7 :

Number of moles and number of molecules in 1.4187 L of $S O_{2}$ at STP respectively are [2026]

Q 8 :

The correct order of total number of atoms present in

(A) 2 moles of cyclohexane

(B) 684 g of sucrose

(C) 90.8 L of dihydrogen at STP

is: [2026]

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	List I		List II
A.	16 g of ${CH}_{4} (g)$	I.	Weighs 28 g
B.	1 g of $H_{2} (g)$	II.	$60.2 \times 10^{23}$ electrons
C.	1 mole of $N_{2} (g)$	III.	Weighs 32 g
D.	0.5 mol of ${SO}_{2} (g)$	IV.	Occupies 11.4 L volume at STP

	List – I		List – II
	(Mass of substance)		(Number of atoms)
A.	1.8 mg water	I.	$2 \times 10^{- 4} \times N_{A}$
B.	9.8 mg sulphuric acid	II.	$1.5 \times 10^{- 4} \times N_{A}$
C.	1.8 mg carbon	III.	$3 \times 10^{- 4} \times N_{A}$
D.	5.85 mg salt (NaCl)	IV.	$7 \times 10^{- 4} \times N_{A}$

Topic Question Set

Q 1 : 0.05 cm thick coating of silver is deposited on a plate of 0.05 m2 area. The number of silver atoms deposited on plate are _________ ×1023. (At mass Ag = 108, d = 7.9 g cm-3) [2024]

Q 2 : 2.8×10-3 mol of CO2 is left after removing 1021molecules from its ‘x’ mg sample. The mass of CO2 taken initially is Given: NA=6.02×1023 mol-1 [2025]

Q 3 : Among 10-9 g (each) of the following elements, which one will have the highest number of atoms? Element: Pb, Po, Pr and Pt [2025]

Q 4 : The number of molecules and moles in 2.8375 litres of O2 at STP are respectively [2023]

Q 5 : Match List I with List II. [2023] List I List II A. 16 g of CH4(g) I. Weighs 28 g B. 1 g of H2(g) II. 60.2×1023 electrons C. 1 mole of N2(g) III. Weighs 32 g D. 0.5 mol of SO2(g) IV. Occupies 11.4 L volume at STP Choose the correct answer from the options given below:

Q 6 : In the reaction, 2Al(s)+6HCl(aq)→2Al3+(aq)+6Cl-(aq)+3H2(g) [2026]

Q 7 : Number of moles and number of molecules in 1.4187 L of SO2 at STP respectively are [2026]

Q 8 : The correct order of total number of atoms present in (A) 2 moles of cyclohexane (B) 684 g of sucrose (C) 90.8 L of dihydrogen at STP is: [2026]

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Q 1 :

0.05 cm thick coating of silver is deposited on a plate of 0.05 $m^{2}$ area. The number of silver atoms deposited on plate are _________ $\times 10^{23}$ . (At mass Ag = 108, d = 7.9 g $c m^{- 3}$ ) [2024]

Q 2 :

$2.8 \times 10^{- 3}$ mol of ${CO}_{2}$ is left after removing $10^{21}$ molecules from its $‘ x ’$ mg sample. The mass of ${CO}_{2}$ taken initially is

Given: $N_{A} = 6.02 \times 10^{23} {mol}^{- 1}$ [2025]

Q 3 :

Among $10^{- 9}$ g (each) of the following elements, which one will have the highest number of atoms?

Element: Pb, Po, Pr and Pt [2025]

Q 4 :

The number of molecules and moles in $2.8375 litres$ of $O_{2}$ at STP are respectively [2023]

Q 6 :

In the reaction,

$2 Al (s) + 6 HCl (aq) \to 2 {Al}^{3 +} (aq) + 6 {Cl}^{-} (aq) + 3 H_{2} (g)$ [2026]

Q 7 :

Number of moles and number of molecules in 1.4187 L of $S O_{2}$ at STP respectively are [2026]

Q 8 :

The correct order of total number of atoms present in

(A) 2 moles of cyclohexane

(B) 684 g of sucrose

(C) 90.8 L of dihydrogen at STP

is: [2026]