One mole of any substance contains `6.022xx10^(23)` atoms/molecules. Number of molecules of `H_(2)SO_(4)` present in 100mL of 0.02M `H_2SO_(4)` solution is

Options

A. `12.044xx10^(20)` molecules
B. `6.022xx10^(23)` molecuels
C. `1xx10^(23)` molecuels
D. `12.044xx10^(23)` molecules

Explanation

One mole of any substance contains `6.022xx10^(23)` atoms/molecuels.
Hence, Number of millimoles of `H_(2)SO_(4)`
=molarity xx volume in mL
`=0.02xx100=2` millimoles
`=2xx10^(-3)` mol
Number of molecuels =number of moles `xxN_(A)`
`=2xx10^(-3)xx6.022xx10^(23)`
`=12.044x10^(20)` molecuels