If `3.01xx10^(20)` molecules of `H_(2)SO_(4)` are removed from 98mg of `H_(2)SO_(4)`. Then number of moles of `H_(2)SO_(4)` left are:

Options

A. `0.1xx10^(-3)` mol
B. `0.5xx10^(-3)` mol
C. `1.66xx10^(-3)` mol
D. `9.95xx10^(-2)` mol

Answer

B

Explanation

Removed number of moles of `H_(2)SO_(4)=(3.01xx10^(20))/(6.02xx10^(23))`
`=5xx10^(-4)`
`=0.5xx10^(-3)`
Initial number of moles of `H_(2)SO_(4)=(0.098)/(98)=0.001=10^(-3)`
Remaining moles of `H_(2)SO_(4)=10^(-3)-0.5xx10^(-3)=0.5xx10^(-3)`