Réponse :
Bonjour,
a) On sait que:
V₀ = 2 mL = 2 × 10⁻³ L
C₀ = 3 × 10⁻² g.L⁻¹
C₁ = 6 × 10⁻³ g.L⁻¹
On cherche:
V₁ = ?
V₀ × C₀ = V₁ × C₁
⇔ V₁ = (V₀ × C₀)/C₁
⇔ V₁ = (2 × 10⁻³ × 3 × 10⁻²)/6 x 10⁻³
⇔ V₁ = 1 × 10⁻² L
b) Il faudra prélever 2,0 mL d'eau de Dakin.
c) 60 µg = 6 × 10¹ × 10⁻⁶ = 6 × 10⁻⁵ g
80 µg = 8 × 10¹ × 10⁻⁶ = 8 × 10⁵ g
110 µg = 1,1 × 10² × 10⁻⁶ = 1,1 × 10⁻⁴ g
130 µg = 1,3 × 10² × 10⁻⁶ = 1,3 × 10⁻⁴ g
[tex]C_{m} S_{1} =\dfrac{mS_{1}}{v}=\dfrac{6*10^-5}{1*10^{-2}} = 6*10^{-3} g.L^{-1}\\\\C_{m}S_{2} = \dfrac{mS_{2}}{v}= \dfrac{8*10^-5}{1*10^{-2}} = 8*10^{-3}g.L^{-1}\\\\C_{m}S_{3} = \dfrac{mS_{3}}{v} = \dfrac{1,1*10^{-4}}{1*10^{-2}}=1,1*10^{-2}g.L^{-1}\\\\C_{m}S_{4} = \dfrac{mS_{4}}{v} = \dfrac{1,3*10^{-4}}{1*10^{-2}}=1,3*10^{-2}g.L^{-1}\\\\[/tex]
d)
[tex]C_{m}_{Dakin}=\dfrac{m}{v} = \dfrac{1}{100} = 1*10^{-2} mg.mL^{-1} = 1*10^{-2} g.L^{-1}[/tex]
8 × 10⁻³ g.L⁻¹ < 1 × 10⁻² g.L⁻¹ < 1,1 × 10⁻² g.L⁻¹
[tex]C_mS_2 < C{mDakin} < C_mS_3[/tex]
