Question

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
A. `12.044xx10^20`
B. `6.022xx10^23`
C. `1xx10^23`
D. `12.044xx10^23`

Answer 1

Molarity `=("No. of moles of solute" xx 1000)/("Volume of solution (in mL)")`
`therefore " 1M " H_2SO_4` contains 1 mole of `H_2SO_4` in 1000 mL of solution
`therefore " 0.02M " H_2SO_4` in 100 mL of solution contains
`(0.02xx100)/(1000)=2xx10^(-3)` mol
`=2xx10^(-3) xx 6.022xx10^23=12.044xx10^20` molecules