Match List-I with List-II List-I List-II A $16g$ of ${\mathrm{CH}}_{4}(g)$ I Weighs $28g$ B $1g$ of ${H}_{2}(g)$ II $60.2\times {10}^{23}$ electrons C $1\mathrm{mole}$ of ${N}_{2}(g)$ III Weighs $32g$ D $0.5\mathrm{mol}$of ${\mathrm{SO}}_{2}(g)$ IV Occupies$11.4L$ volume at STP Choose the correct answer from the options given below:

Match List-I with List-II List-I List-II… - JEE Main 2023 Chemistry PYQ

JEE Main 2023 — Chemistry Physical Chemistry

2023mcqeasy

Match List-I with List-II

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

Choose the correct answer from the options given below:

Official previous-year question

Held on 10 Apr 2023 · Verified 6 Jul 2026.

Options

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

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Solution

The number of moles $=\frac{\mathrm{Mass}}{\mathrm{Molar}\mathrm{mass}}$

The number of molecules $=\mathrm{moles}\times {N}_{A}$

One mole of any gas occupies $22.7L$ at STP.

(A) Moles of ${\mathrm{CH}}_{4}=\frac{16}{16}=1$

Molecules $=6.02\times {10}^{23}$

Electrons $=10\times 6.02\times {10}^{23}$

$=60.2\times {10}^{23}$

(B) Moles of ${H}_{2}=\frac{1}{2}=0.5$ mole

Volume at STP $=\frac{1}{2}\times 22.7=11.35\mathrm{lit}$ .

(D) $0.5$ mole ${\mathrm{SO}}_{2}=\frac{1}{2}\times 64=32\mathrm{gm}$

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