n=Q/F=965/96500=0.01000mol n(1/2Cu)=0.2859×2/63.54=0.008999mol n(H)=n-n(Cu)=0.01000mol-0.008999mol =0.00100mol V(H2)= n(H,)RT p 0-00100×8-314×298-15m3=0-0122dm3 2×101.325 3.电解质溶液的导电性质 1)电导、电导率和摩尔电导率 (1)定义 电导即电阻的倒数，用符号G表示，即 G=1/R (6-1-1)n = Q / F = 965 / 96500 = 0.01000 mol n（1/2Cu）= 0.2859×2/63.54 = 0.008999 mol n（H）= n － n（Cu）= 0.01000 mol －0.008999 mol = 0.00100 mol ( ) p n RT V 2 2 H (H ) = 3 3 m 0 0122dm 2 101 325 0 00100 8 314 298 15 = ⋅ × ⋅ ⋅ × ⋅ × ⋅ = 3. 电解质溶液的导电性质 1）电导、电导率和摩尔电导率 （1）定义 电导即电阻的倒数，用符号G 表示，即 G = 1 / R （6-1-1）