Calculate the molarity of each of the following solutions :
30 mL of 0.5 M `H_(2)SO_(4)` diluted to 500 mL.

Explanation

Number of moles present in 1000 mL 0.5 M `H_(2)SO_(4)=0.5` mol
`therefore` Number of moles present in 30 mL of 0.5 M
`H_(2)SO_(4)=(0.5xx30)/(1000)` mol
= 0.015 mol
Therefore, molarity `= (0.015)/(0.5L)` mol
= 0.03 mol.