`a)` `m_{\text{dd}}=d.V=1,28.250=320g`
`=> m_{\text{ct}}=\frac{320.20%}{100%}=64g`
`=> n_{NaOH}=\frac{m_{\text{ct}}}{M}=\frac{64}{40}=1,6(mol)`
`b)` `m_{\text{dd}}=d.V=1,83.150=274,5g`
`=> m_{\text{ct}}=\frac{274,5.93,6%}{100%}=256,932g`
`=> n_{H_2SO_4}=\frac{m_{\text{ct}}}{M}=\frac{256,932}{98}\approx 2,62(mol)`