m\(_{H_2O}\) = D*V = 1*500 = 500 (gam)
n\(_{Na_2O}\) = \(\dfrac{0,2}{62}\) \(\approx\) 0,03 (mol)
PTHH :
Na\(_2\)O + H\(_2\)O → 2NaOH
mol 0,03 → 0,06
Dung dịch thu được là NaOH
m\(_{ddsaupu}\) = m\(_{Na_2O}\) + m\(_{H_2O}\) = 0,2 + 500 = 500,2 (gam)
Vậy:
C%\(_{NaOH}\) = \(\dfrac{m_{NaOH}\cdot100\%}{m_{ddsaupu}}\) = \(\dfrac{0,06\cdot40\cdot100}{500,2}\) \(\approx\) 0,48(%)