1.

The molar conductivity of a 0.5 mol/dm^(3) solution of AgNO_(3) with electrolytic conductivity of 5.76xx10^(-3)" S "cm^(-1) at 298 K is

Answer»

`2.88" S "CM^(2)//mol`
`11.52" S "cm^(2)//mol`
`0.086" S "cm^(2)//mol`
`28.8" S "cm^(2)//mol`

Solution :`wedge_(m)=(kappaxx1000)/("Molarity")`
`=((5.76xx10^(-3)" S "cm^(-1))(1000cm^(3)L^(-1)))/(0.5" mol "L^(-1))`
`=(5.76)/(0.5)=11.52" S "cm^(2)mol^(-1)`.


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