Question

Answer 1

$m_{dd\ NaOH} = D.V = 1,5.160 = 240(gam)$
$n_{NaOH} = \dfrac{240.25\%}{40} = 1,5(mol)$
$2NaOH + H_2SO_4 \to Na_2SO_4 + 2H_2O$

$n_{H_2SO_4} = \dfrac{1}{2}n_{NaOH} = 0,75(mol)$
$V_{dd\ H_2SO_4} = \dfrac{0,75}{2,5} = 0,3(lít)$

Answer 2

\(m_{NaOH}=160.1,5=240\left(g\right)\)

\(n_{NaOH}=\dfrac{240.25\%}{40}=1,5\left(mol\right)\)

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

  1,5               0,75                                   ( mol )

\(V_{H_2SO_4}=\dfrac{0,75}{2,5}=0,3\left(l\right)\)