Question

What is the total number of atoms present in 25.0 mg of camphor, `C_(10)H_(16)O`?
A. `9.89xx10^(19)`
B. `6.02xx10^(20)`
C. `9.89xx10^(20)`
D. `2.67xx10^(21)`

Answer 1

Correct Answer - D
Molar mass of `C_(10)H_(16)O = 12+16+16 = 152 g mol^(-1)`
`25.0 mg = 25.0 mgxx((1)/(10^(3)mg))xx((1mol)/(152g))xx(27xx6.02xx10^(23)atoms)/(1mol)`
`=(25xx27xx6.02)/(152)xx10^(20)` atom
`= 26.73xx10^(20)` atom `= 2.67xx10^(21)` atoms