A weak monobasic acid is 5% dissociated in 0.01 mol `dm^(-3)` solution. The limiting molar conductivity at infinite dilution is `4.00xx10^(-2) ohm^(-1) m^(2) mol^(-1)`. Calculate the conductivity of a 0.05 mol `dm^(-3)` solution of the acid.

Options

A. `8.94xx10^(-6) "ohm"^(-1) cm^(2) mol^(-1)`
B. `8.92xx10^(-4) "ohm"^(-1) cm^(2) mol^(-1)`
C. `4.46xx10^(-6) "ohm"^(-1) cm^(2) mol^(-1)`
D. `2.23xx10^(-5) "ohm"^(-1) cm^(2) mol^(-1)`

Answer

B

Explanation

`K_a=Calpha^2=0.01xx(0.05)^2=2.5xx10^(-5)`
`K_a=Calpha^2`
`2.5xx10^(-5)=0.05xxalpha^2 rArr alpha = 0.0223`
`alpha=Lambda_m^c/Lambda_m^oo`
`Lambda_m^c = 0.0223 xx 4 xx 10^(-2) = 8.92 xx 10^(-4) "ohm"^(-1) "cm"^(2) "mol"^(-1)`