Match List I with List II. Weighs 28 g [2023]

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Q.
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:

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

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

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

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

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

Solution

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

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

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

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

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

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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

Solution

Q. 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:

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

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

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

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

Ans. (4) (A) 16 g or 1 mol CH4 or 6.02×1023×10 electrons =60.2×1023 electrons (II) (B) 1 g H2 or 12 mol H2 or 11.35 L volume at STP (IV) (C) 1 mol N2=28 g N2 (I) (D) 0.5 mol SO2=32 g SO2 (III)

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