\(n_{H_2SO_4}=0.45mol\)

a. \(H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\)

      0.45          0.9

\(m_{dd_{NaOH}}=\dfrac{\left(0.9\times40\times100\right)}{40}=90g\)

b. \(H_2SO_4+2KOH\rightarrow K_2SO_4+2H_2O\)

      0.45          0.9

\(V_{KOH}=\dfrac{n}{C_M}=\dfrac{0.9}{2}=0.45l=450ml\)

\(D_{KOH}=\dfrac{m_{KOH}}{V_{KOH}}\Leftrightarrow m=1.045\times450=470.25g\)