1.

The limiting molar conductivities ^^ for NaCl, KBr and KCl are 126,152 and 150 S cm^(2)mol^(-1) respectively. The ^^ for NaBr is

Answer»

278 S `cm^(2)mol^(-1)`
176 S `cm^(2)mol^(-1)`
128S `cm^(2)mol^(-1)`
302 S `cm^(2)mol^(-1)`

Solution :`(126s" "cm^(2))wedge_(NaCl)^(o)=wedge_(Na^(+))^(o)+wedge_(Cl^(-))^(o)` . . .(i)
`(152s" "cm^(2))wedge_(KBr)^(o)=wedge_(K^(+))^(o)+wedge_(Br^(-))^(o)` . . . (II)
`(150s" "cm^(2))wedge_(KCL)^(o)=wedge_(K^(+))^(o)+wedge_(Cl^(-))^(o)` . . .(iii)
by EQUATION (i)+(ii)-(iii)
`because wedge_(NABR)^(o)=wedge_(Na^(+))^(o)+wedge_(Br^(-))^(o)` ltBrgt `=126+152-150=128s" "cm^(2)mol^(-1)`.


Discussion

No Comment Found

Related InterviewSolutions