If `3*01xx10^(20)` molecules are removed from 98 mg of `H_(2)SO_(4)`, then the number of moles of `H_(2)SO_(4)` left are

Options

A. `0.5 xx 10^(-3)` mol
B. `0.1 xx 10^(-3)` mol
C. `9.95 xx 10^(-2)` mol
D. `1.66 xx 10^(-3)` mol

Answer

A

Explanation

No. of moles of `H_(2)SO_(4)` in 98 mg `=(98 xx 10^(-3))/98 = 1 xx 10^(-3)`
No. of moles of `3.01 xx 10^(20)` molecules of
`H_(2)SO_(4) = (3.01 xx 10^(20))/(6.022 xx 10^(23))`
`=0.5 xx 10^(-3)`
Moles of `H_(2)SO_(4)` left `=1 xx 10^(-3) = 0.5 xx 10^(-3)`