Molar conductance values at infinite dilution of Na+ and Cl- ions are 50.11*10^-4Sm^2mol^-1 and 76.34*10^-4Sm^2mol^-1 respectively. Calculate the transport number of Na+ ion.

Asked by mrudulmahadev1311 | 22nd Aug, 2019, 07:47: AM

Expert Answer:

Given:
 
Molar conductance at infinite dilution;
 
λ+Na = 50.11×10Sm2mol−1 
 
λ+Cl = 76.34×10Sm2mol−1 
 
Molar conductance of NaCl is = λ+Na + λ+Cl 
 
                                           = 50.11×104 + 76.34×10 
 
                                           = 126.45×104 Sm2mol−1 
 
Transport number of Na+ 
                                 straight t subscript plus equals space end subscript fraction numerator straight lambda subscript 0 superscript plus over denominator straight capital lambda subscript 0 end fraction

space space space space equals fraction numerator 50.11 cross times 10 to the power of negative 4 end exponent over denominator 126.45 cross times 10 to the power of negative 4 end exponent end fraction

space space space space equals 0.3962 
 

Answered by Varsha | 22nd Aug, 2019, 12:32: PM