Vs = 500 mL = 0,5 L , p = 1,14 g/mL , Cm = 4,3 mol/L
metoda I:
Cm = md/MxVs ; M.HNO3 = 1+14+3x16 = 63 g/mol
=> md = CmxMxVs = 4,3x63x0,5 = 135,45 g HNO3 dizolvati
w% = mdx100/ms ; p = ms/Vs
=> ms = pxVs = 1,14x500 = 570 g sol. HNO3
=> w% = 135,45x100/570 = 23,76%
metoda II:
Cm = w%xpx10/M
=> w% = CmxM/10p = 4,3x63/10x1,14 = 23,76%
