Question

What is the total number of atoms present in 25.0 mg of camphor, C₁₀H₁₆O?

पर्याय

• 9.89 × 10¹⁹

• 6.02 × 10²⁰

• 9.89 × 10²⁰

• 2.67 × 10²¹

Answer 1

2.67 × 10²¹

Explanation:

Molar mass of C₁₀H₁₆O = 120 + 16 + 16 = 152 g mol⁻¹

25.0 mg = `25.0  "mg" xx ((1 "g")/(10^3 "mg")) xx ((1 "mol")/(152 "g")) xx ((27 xx 6.02 xx 10^23 "atoms")/(1 "mol"))`

= `(25 xx 27 xx 6.02)/152 xx 10^20  "atom"`

= 2.67 × 10²¹ atoms