0.05 M NaOH solution offered a resistance of `3.1*6Omega` in a conductivity cell at 298 K. If the cell constant of the cell si `0*367cm^(-1),` calculate the molar conductivity of NaOH solution.

Options

A. `234 S cm^(2) mol^(-1)`
B. `23.2 S cm^(2) mol^(-1)`
C. `4645 S cm^(2) mol^(-1)`
D. `5464 S cm^(2) mol^(-1)`

Answer

B

Explanation

Here , R = 31.6 ohm
`therefore` Conductance = `(1)/(R) = (1)/(31.6) ohm^(-1) = 0.0316 ohm^(-1)`
Specific conductance = conductance `xx` cell constant = `0.0316 ohm^(-1) xx 0.367 cm^(-1)`
= `0.0116 ohm^(-1) cm^(-1)`
Now molar concentration = 0.5 M (given) =` 0.5 xx 10^(-3) mol cm^(-3)`
`therefore` Molar conductance = `(k)/(c) = (0.0116)/(0.5 xx 10^(-3))`
`= 23.2 S cm^(2) mol^(-1)`