m = 8g NaOH
V = 500 cm³ = 0,5 dm³ = 0,5 L
concentratia molara(CM) = numar de moli/volum
numar de moli(M) = m/n
n NaOH = 23+16+1 = 40 g/mol
M = 8/40 = 1/5
CM = [tex] \frac{ \frac{1}{5} }{0,5} = \frac{ \frac{1}{5} }{ \frac{1}{2} } = \frac{2}{5} = 0,4 [/tex] g/L