Question

01. Identify the option containing maximum
number of atoms.
(A) 18 mg of glucose
(B)2 mg of hydrogen gas
(C) 10 mg of H20
(D) 7.8 mg of benzene​

Answer 1

Answer:

hence 10 mg hydrogen contains maximum number of atoms.

Answer 2

Given: (A) 18 mg of glucose

(B)2 mg of hydrogen gas

(C) 10 mg of H20

(D) 7.8 mg of benzene

To find: We have to find the compound which contains the maximum number of atoms.

Solution:

(A)

180 grams of glucose contains $6.023×10^{23}$ number of atoms.

Thus 18mg or $18×10^{-3}$ grams of glucose will contain

$6.023 \times {10}^{ 23} \times 18 \times {10}^{ - 3} \times \frac{1}{180} \\ = 6.023 \times {10}^{19}$

(B)

2grams of hydrogen contains $6.023×10^{23}$ number of atoms.

Thus 2mg or $2×10^{-3}$ grams of hydrogen gas will contain

$6.023 \times {10}^{ 23} \times 2 \times {10}^{ - 3} \times \frac{1}{2} \\ = 6.023 \times {10}^{20}$ atoms.

(C)

18grams of water contains $6.023×10^{23}$ number of atoms.

Thus 10mg or $10×10^{-3}$ grams of water will contain

$6.023 \times {10}^{ 23} \times 10 \times {10}^{ - 3} \times \frac{1}{18} \\ = 3.3 \times {10}^{20}$ atoms.

(D)

78grams of benzene contains $6.023×10^{23}$ number of atoms.

Thus 7. 8mg or $7.8×10^{-3}$ grams of benzene will contain

$6.023 \times {10}^{ 23} \times 7.8 \times {10}^{ - 3} \times \frac{1}{78} \\ = 0.6023 \times {10}^{20}$ atoms.

Thus 2mg hydrogen gas will contain the maximum number of atoms.

The correct option is B.