At `18^(@)C`, the conductivities at infinite dilution of `NH_(4)Cl, NaOH` and `NaCl` are 129.8, 217.4 and 108.9 mho respectively. If the equivalent conductivity of N/100 solution of `NH_(4)OH` is 9.93 mho, calculate the degree of dissociation of `NH_(4)OH` at this dilution.

Options

A. 0.392
B. 39.2
C. 3.92
D. 0.039

Answer

C

Explanation

` ^^_(NaOH)^(0)=^^_(NH_(4))^(0)+ +lambda_(Cl^(-))^(0)=129.8 " "...(1)`
`^^_(NaOH)^(0)=^^_(Na)^(0)++lambda_(OH^(-))^(0)=217.4" "...(2)`
`^^_(NaCl)^(0)=^^_(Na)^(0)++lambda_(Cl^(-))^(0)=108.9" "...(3)`
`^^_(NH_(4)OH)^(0)=^^_(NH_(4))^(0)++lambda_(OH^(-))^(0)=129.8`
Adding (1), (2) and subtracting (3),
`^^_(NH_(4)OH)^(0)=238.3Omega^(-1) "cm"^(2) "mol"^(-1)`
`alpha=(^^_(m)^(c))/(^^_(m)^(c))=(9.33)/(238.3)=0.03915`
% dissocation `=0.03915 xx 100=3.915 %`