What is the total number of atoms present in 25.0 mg of camphor, `C_(10)H_(16)O`?

Options

A. `2.57xx10^(21)`
B. `9.89xx10^(19)`
C. `2.67xx10^(21)`
D. `6.02xx10^(20)`

Answer

C

Explanation

Molecular mass of camphor
`=10xx12+16xx1+1xx16`
`=120+16+16=152`
No. of mass `=("Given mass")/("Molecular mass")=(25xx10^(-3))/(152)`
`=1.64xx10^(-4)`
Number of atoms in 1 mole of camphor `(C_(10)H_(16)O)`
`=10+16++1=7xx6.022xx10^(23)`
Thus, no. of atoms in `1.64 xx 10^(-4)` moles will be
`6.022xx27xx1.64xx10^(-4)xx10^(23)=2.67xx10^(21)`