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?

Options

A. `147 "ohm"^(-1)cm^2"mole"^(-1)`
B. `107 "ohm"^(-1)cm^2"mole"^(-1)`
C. `127 "ohm"^(-1)cm^2"mole"^(-1)`
D. `157 "ohm"^(-1)cm^2"mole"^(-1)`

Answer

C

Explanation

`lambda_m^C=lambda_m^oo-bsqrtC`
when `C_1=4xx10^(-4) lambda_m^C=107`
and when `C_2=9xx10^(-4) lambda_m=97`
so `107=lambda_m^oo-bxx2xx10^(-2)`...(1)
`97=lambda_m^oo - bxx3xx10^(-2)`...(2)
b=1000
`lambda_m=lambda_m^oo-bsqrtC`
`lambda_m^oo=lambda_m+bsqrtC`
`=107xx10^3xx2xx10^(-2)`
`lambda_m^oo=127 "ohm"^(-1)cm^(2)"mole"^(-1)`