Question

The limiting molar conductivities `^^` for NaCl, KBr and KCl are 126,152 and 150 S `cm^(2)mol^(-1)` respectively. The `^^` for NaBr is
A. 278 S `cm^(2)mol^(-1)`
B. 176 S `cm^(2)mol^(-1)`
C. 128S `cm^(2)mol^(-1)`
D. 302 S `cm^(2)mol^(-1)`

Answer 1

Correct Answer - C
`(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)`.