1.

0.05 M NaOH solution offered a resistance of 31.6 ohm in a conductivity cell. If the cell constant of cell is 0.367cm^(-1), calculate the molar conductivity of the solution.

Answer»

Solution :MOLARITY =0.05
`R=31.6Omega`
Cell consatant, `(L)/(a)=0.367cm^(-1)`
MOLAR conductivity,
`Delta_(m)=(Kxx1000)/("Molarity")`
`=((1)/(R)xx(l)/(a)xx100)/("Molarity")`
`=(1)/(31.6)xx(0.367xx1000)/(0.5)`
`=(0.367xx1000)/(31.6xx0.05)`
`=232.2ohm^(-1)cm^(2)MOL^(-1)`


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