Se requiere preparar 150 ml de solucion al 15 % de acido nitrico deluido y de densidax 1.11 g/ml .determine la concentracion N
lqe125
Acido nítrico: HNO3 D = m/V 1.11 = m/150 mtotal = 166.5 g m = 0.15*166.5 = 24.975 g Masa molar del HNO3 = 1+14+16*3 = 63 g 24.975g*(1mol/63g) = 0.396 moles n = M*V 0.396 = M*0.15 M = 2.64 N = M*θ N = 2.64*1 = 2.64
D = m/V
1.11 = m/150
mtotal = 166.5 g
m = 0.15*166.5 = 24.975 g
Masa molar del HNO3 = 1+14+16*3 = 63 g
24.975g*(1mol/63g) = 0.396 moles
n = M*V
0.396 = M*0.15
M = 2.64
N = M*θ
N = 2.64*1 = 2.64