Question

What is the total number of atoms present in 25.0 mg of
camphor, CH.0?
a. 9.89 x 1019 b. 6.02 x 1020
c. 9.89 x 1020 d. 2.67 x 1021
a​

Answer 1

Answer:

2.67*10 raise to 21

Explanation:

Answer 2

Answer:

Chemical formula of camphor is C₁₀H₁₆O

Molar mass of camphor = 152 gm

∴ Moles of camphor in 25 mg =  [25 × 10⁻³ ]÷ 152

∴ Total atoms = 6.02 × 10²³ × moles × 27

                        =    [25 × 10⁻³ ] ÷ 152 }× 6.02 × 10²³  × 27

∴ There are 2.67 x 10²¹atoms