$n_{Na}=2,3/23=0,1mol$
$a.2Na+2H_2O\to 2NaOH+H_2↑$
b.Theo pt :
$n_{NaOH}=n_{Na}=0,1mol$
$⇒m_{NaOH}=0,1.40=4g$
c.Theo pt :
$n_{H_2}=1/2.n_{Na}=1/2.0,1=0,05mol$
$m_{ddspu}=2,3+36,7-0,05.2=38,9g$
$⇒C\%_{NaOH}=\dfrac{4}{38,9}.100\%=10,28\%$
$d.V_{NaOH}=\dfrac{m_{dd}}{d}=\dfrac{38,9}{1,01}=38,51ml=0,03851l$
$⇒C_{M_{NaOH}}=\dfrac{0,1}{0,03851}=2,6(mol/l)$