Question

When $0.1$mol $CoCl_{3}(NH_{3})_{5}$is treated with excess of $AgNO_{3}0.2mol$of AgCl are obtained. The conductivity of solutions will correspond to

• A. $1:3$electrolyte
• B. $1:2$electrolyte
• C. $1:1$electrolyte
• D. $3:1$electrolyte

Answer 1

Answer: (B)

$1:2$electrolyte

Answer 2

$0.2$moles of AgCl are obtained when $0.1$mole $CoCl_{3}(NH_{3})_{5}$is treated with excess of $AgNO_{3}$which shows that one molecules of the complex given two $Cl^{-}$ion in solutions. Thus, the formula of the complex is $\lbrack Co(NH_{3})_{5}Cl_{2}\rbrack$ i.e., $1:2$electrolyte.