1.

A 0.05M KOHsolution offered resistanceof 31.6 omegain a conductivitycell of cel constant 0.3967 cm^(-1)at 298 k what is the molar conductance of KOHsolution

Answer»

150.3
18068
232
215.7

Solution :Givencellconstant `=0.367 cm^(-1)`
Resistance =31.6 `omega` and KOHis monoacidic base so itsmolarity =0.05 M
k(specific conuctance ) `= "CONDUCTANCE" xx "cell CONSTANT"`
`=(1)/("resistance")xx("cell constant")`
`=(0.367 cm^(-1))/(31.6 omega)=0.0116 omega^(-1) cm^(-1)`
`=0.0116 S CMS^(-1)`
`therefore` (molar conductance )`=(kxx1000)/(M)`
`=(kxx1000)/(0.05)=(0.0116xx1000)/(0.05)`
`=232 S cm^(2) mol^(-1)`


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