Question

What is the volume of 3.00 M sulfuric aid that contain 9.809 g of H2SO4 (98.09g/mol)

Answer 1

Given :

Molarity of sulfuric acid solution is 3.0 M.

Amount of sulfuric acid present in solution is 9.809 g.

To Find :

The volume of solution.

Solution :

We know, molarity is given by :

$Molarity = \dfrac{number \ of \ moles \times 1000}{Volume\ ( ml )}\\\\M = \dfrac{w \times 1000}{M.M \times V}\\\\3 = \dfrac{9.809\times 1000}{98.09 \times V}\\\\V = \dfrac{1000}{10\times 3}\ ml\\\\V = 33.33 \ ml$

Therefore, volume required is 33.33 ml .