One mole of any substance contains `6.022 xx 10^(23)` atoms/molecules. Number of molecules of `H_(2)SO_(4)` present in 100 mL of 0.02 M `H_(2)SO_(4)` solution is :

Options

A. `12.044 xx 10^(20)` molecules
B. `6.022 xx 10^(23)` molecules
C. `1xx 10 ^(23)` molecules
D. `12.044 xx 10^(23)` molecules

Answer

A

Explanation

One mole of any substance contains `6.022 xx 10^(23)` atoms/molecules.
Hence, Number of millimoles of `H_(2) SO_(4)`
= molarity `xx` volume in mL
` = 0.02 xx 100` = 2 millimoles
` = 2 xx 10^(-3)` mol
Number of molecules = number of moles `xx N_(A)`
` = 2 xx 10^(-3) xx 6.022 xx 10^(23)`
` = 12.044 xx 10^(20)` molecules.