1.

The molarconductivityat zeroconcentrationofNH_(4) CI NaOHand NaCIare respectively 149.7Omega^(-1) cm^(2) mol^(-1) 248.1 Omega^(-1) cm^(2) mol^(-1)and 126.5 Omega^(-1) cm^(2) mol^(-1). Whatis themolarconductivityof NH_(4)OH atzeroconcentration ?

Answer»


SOLUTION :`^^_(o_((NH_(4)C1)) = 149.1 Omega^(-1) CM^(2) mol^(-1)`
`^^_(o_((NAOH)))=248.1 Omega^(-1) cm^(2) mol^(-1)`
`^^_(o_((NaCI)))=126.5 Omega^(-1) cm^(2) mol^(-1)`
By Kohlrausch's law
`^^_(o_((NH_(4)OH)))= lambda_(NH_(4))^(0) + lambda_(OH^(-))^(0)""......(I)`
`^^_(o_((NH_(4)C1)))=lambda_(NH_(4))^(0) + lambda_(C1)^(0)""......(i)`
`^^_(o_((NaOH)))=lambda_(Na^(+))^(0) + lambda_(OH^(-))^(0) ""......(ii)`
`^^_(0_((NaC1)))=lambda_(Na^(+))^(0) + LAMBDA _(C1^(-))^(0)""......(iii)`
Addingequations(i) and (ii) and subtractingequation (iii) wegetequation I.
`:. ^^_(o_((NH_(4)OH))) = ^^_(o_((NH_(4)C1))) + ^^_(0_((NaOH))) ^(-) ^^_(0_((NaC1)))`
`=149.7 + 248 .1- 126. 5`
`=271.3 Omega^(-1)cm^(2) mol^(-1)`


Discussion

No Comment Found