The given concentration of boric acid = 0.0500 M
Required volume of the solution = 2 L
Molarity is the moles of solute present per liter solution. So 0.0500 M boric acid has 0.0500 mol boric acid present in 1 L solution.
Calculating the moles of 0.0500 M boric acid present in 2 L solution:
[tex]2 L * \frac{0.0500 mol B(OH)_{3} }{1 L} = 0.100 mol B(OH)_{3}[/tex]
Converting moles of boric acid to mass:
[tex]0.100 mol B(OH)_{3} * \frac{61.83 g}{mol B(OH)_{3}} = 6.183 g[/tex]
Therefore, 6.183 g boric acid when dissolved and made up to 2 L with distilled water gives 0.0500 M solution.
