\[\kappa = \frac{1}{\rho} = \frac{K_\text{cell}}{R},\qquad \Lambda_m = \frac{\kappa}{c}\]
含义Meaning
电导率 κ 衡量溶液导电能力；除以浓度得摩尔电导率 Λ_m，便于比较不同电解质。强电解质 Λ_m 随浓度增加缓慢下降，弱电解质则急剧下降。
Conductivity κ measures how well a solution conducts; dividing by concentration gives molar conductivity Λ_m for comparing electrolytes. It falls slowly with concentration for strong electrolytes and steeply for weak ones.
符号说明Symbols
| 符号Symbol | 含义 Meaning | 单位 Unit |
|---|---|---|
| \(\kappa\) | 电导率conductivity | S·m⁻¹ |
| \(K_\text{cell} = l/A\) | 电导池常数（用 KCl 标准溶液标定）cell constant (calibrated with KCl) | m⁻¹ |
| \(\Lambda_m\) | 摩尔电导率molar conductivity | S·m²·mol⁻¹ |
| \(c\) | 浓度（mol/m³！）concentration (mol/m³!) | mol·m⁻³ |
推导Derivation
溶液电阻 \(R = \rho l/A\)，电导 \(G = 1/R = \kappa A/l\)。
R = ρl/A, so G = κA/l.
κ 正比于载流子浓度，所以除以 c 得到每摩尔电解质的贡献。
κ scales with carrier concentration; divide by c.
适用条件与局限Assumptions & limitations
使用交流电避免电极极化；c 用 mol/m³ 时 Λ_m 的单位为 S·m²/mol（常用 S·cm²/mol 需换算）。
Use AC to avoid polarisation; watch units.
例题Worked example
0.0100 mol/L KCl 在 25 °C 的 κ = 0.1413 S/m：Λ_m = 0.1413/10 = 0.01413 S·m²/mol = 141.3 S·cm²/mol。
0.0100 M KCl: Λ_m = 141.3 S cm² mol⁻¹.