Solution :
Formule : \(\sigma = \sum \lambda_i \times [X_i]\)
Pour NaCl : \(\sigma = (\lambda_{\text{Na}^+} + \lambda_{\text{Cl}^-}) \times C\)
\(1,24 = (5,0 + 7,6) \times 10^{-3} \times C\)
\(C = \frac{1,24}{12,6 \times 10^{-3}} = 98,4 \, \text{mol.m}^{-3} = 0,0984 \, \text{mol.L}^{-1}\)

Solution :
\(\sigma = (\lambda_{\text{K}^+} + \lambda_{\text{Cl}^-}) \times C\)
\(0,141 = (\lambda_{\text{K}^+} + 7,6 \times 10^{-3}) \times 10\) (car \(0,01 \, \text{mol.L}^{-1} = 10 \, \text{mol.m}^{-3}\))
\(\lambda_{\text{K}^+} = \frac{0,141}{10} – 7,6 \times 10^{-3} = 6,5 \times 10^{-3} \, \text{S.m}^2.\text{mol}^{-1}\)

Solution :
\(\text{CaCl}_2 \rightarrow \text{Ca}^{2+} + 2\text{Cl}^-\)
\(\sigma = \lambda_{\text{Ca}^{2+}}[\text{Ca}^{2+}] + \lambda_{\text{Cl}^-}[\text{Cl}^-]\)
\([\text{Ca}^{2+}] = 5 \, \text{mol.m}^{-3}\), \([\text{Cl}^-] = 10 \, \text{mol.m}^{-3}\)
\(\sigma = (11,9 \times 5 + 7,6 \times 10) \times 10^{-3} = (59,5 + 76) \times 10^{-3} = 0,1355 \, \text{S.m}^{-1}\)

Solution :
Formule : \(\sigma = k \times G\)
\(\sigma = 125 \times 1,5 \times 10^{-2} = 1,875 \, \text{S.m}^{-1}\)

Solution :
\(\sigma_{\text{mélange}} = \frac{V_1\sigma_1 + V_2\sigma_2}{V_1 + V_2}\)
\(\sigma = \frac{0,1 \times 1,2 + 0,2 \times 0,7}{0,3} = \frac{0,12 + 0,14}{0,3} = 0,867 \, \text{S.m}^{-1}\)

Solution :
\(k = \frac{\sigma}{G} = \frac{0,141}{1,13 \times 10^{-3}} = 124,8 \, \text{m}^{-1}\)

Solution :
\(\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-\)
\(\sigma = \lambda_{\text{H}^+}[\text{H}^+] + \lambda_{\text{A}^-}[\text{A}^-]\)
\([\text{H}^+] = [\text{A}^-] = \alpha C = 0,05 \times 100 = 5 \, \text{mol.m}^{-3}\)
\(\sigma \approx (35 + 4,0) \times 10^{-3} \times 5 = 0,195 \, \text{S.m}^{-1}\) (avec \(\lambda_{\text{H}^+} \approx 35 \times 10^{-3}\))

Solution :
La conductivité est proportionnelle à la concentration :
\(\sigma_2 = \frac{\sigma_1}{5} = \frac{1,5}{5} = 0,3 \, \text{S.m}^{-1}\)

Solution :
Calcul des \(\sum \lambda_i\) :
a) NaCl : \(5,0 + 7,6 = 12,6\)
b) MgCl₂ : \(10,6 + 2 \times 7,6 = 25,8\)
c) AlCl₃ : \(18,3 + 3 \times 7,6 = 41,1\)
Ordre : a) < b) < c)

Solution :
1. \(n = C \times V = 0,02 \times 0,25 = 0,005 \, \text{mol}\)
\(m = n \times M = 0,005 \times 142 = 0,71 \, \text{g}\)
2. \(\text{Na}_2\text{SO}_4 \rightarrow 2\text{Na}^+ + \text{SO}_4^{2-}\)
\([\text{Na}^+] = 2 \times 0,02 = 0,04 \, \text{mol.L}^{-1} = 40 \, \text{mol.m}^{-3}\)
\([\text{SO}_4^{2-}] = 0,02 \, \text{mol.L}^{-1} = 20 \, \text{mol.m}^{-3}\)
3. \(\sigma = (5,0 \times 40 + 16,0 \times 20) \times 10^{-3} = (200 + 320) \times 10^{-3} = 0,52 \, \text{S.m}^{-1}\)