Resistance of `0.2 M` solution of an electrolyte is `50 Omega`. The specific conductance of the solution is ` 1.3 S m^(-1)`. If resistance of the `0.4 M` solution of the same electrolyte is `260 Omega`, its molar conductivity is .

Options

A. `6250 sm^(2) "mol"^(-1)`
B. `6.25 xx 10^(-14) "Sm"^(2) "mol"^(-1)`
C. `625 xx10^(-4) sm^(2) "mol"^(-1)`
D. `62.5 sms^(2) "mol"^(-1)`

Answer

B

Explanation

specific conductance=condcutance x cell constant
1.3 `Sm^(-1)=-1/50` S x cell constant
`therefore` cell constant `=1.3x50cm^(-1)=65 m^(-1)=(65/100)cm^(-1)`
molar conductivity `=(1000xx "consduancex cell constant")/("molarity")`
`=1000/0.4 xx1/60xx65/100`
`=6.25 S cm^(2) "mol"^(-1)`
`=6.25 xx10^(-4) Sm^(-2) "mol"^(-1)`
`=6.25xx10^(-4) "Sm"^(2) "mol"^(-1)`