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The molar conductance of NaCl varies with the concentration as shown in the following table and all values follows the equation. `lambda_(m)^(c)=lambda_(m)^(oo)-bsqrt(C)` where `lambda_(m)^(c)`= molar specific conductance `lambda_(m)^(oo)=` molar specific conductance at infinite dilution C=molar concentration When a certain conductivity cell (C) was filled with 25`xx10^(-4)(M) NaCl` solution, the resistance of the cell was found to be 1000 ohm. At infinite dilution, conductance of `Cl^(-)` and `SO_(4)^(2-)` are `80ohm^(-1) cm^(2) "mole"^(-1)` and `160ohm^(-1) cm^(2) "mole"^(-1)` respectively. What is the molar conductance of NaCl at infinite dilution?A. `147ohm^(-1)cm^(2)"mole"^(-1)`B. `107ohm^(-1)cm^(2)s"mole"^(-1)`C. `127ohm^(-1)cm^(2)"mole"^(-1)`D. `157ohm^(-1)cm^(2)"mole"^(-1)` |
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Answer» Correct Answer - C `lamda_(m)^(C)=lamda_(m)^(infty)-bsqrt(C)` when `C_(1)=4xx10^(-4)lamda_(m)^(C)=107` and when `C_(2)=9xx10^(-4),lamda_(m)=97` so `107=lamda_(m)^(infty)-bxx2xx10^(-2)` .(1) `97=lamda_(m)^(infty)-bxx3xx10^(-2)` ..(2) `b=1000` `lamda_(m)=lamda_(m)^(infty)-bsqrt(C)` `lamda_(m)^(infty)=lamda_(m)-bsqrt(C)=107+10^(3)xx2xx10^(-2)` `lamda_(m)^(infty)=127ohm^(-1)cm^(2)"mole"^(-1)` |
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