Question

$20ml$ of $AgNO_{3}$ solution was titrated with $0.20molL^{- 1}$ $KCl$ solution in the conductivity cell. The data were plotted as follows. The moles of $AgNO_{3}$ in the solution initially was

• A. $2\times 10^{- 3}$
• B. $4\times 10^{- 3}$
• C. $8\times 10^{- 3}$
• D. $2\times 10^{- 2}$

Answer 1

Answer: (B)

At equivalence point
Moles of $\text{AgNO}_{3}=Moles$ of $KCl$
$10^{- 3}\times 20\times M=0.20\times 20\times 10^{- 3}$
$M=0.20$
$Moles$ of $AgNO_{3}$ initially $=M\times V$
$=\frac{0 . 20}{100}\times 20\times 10^{- 3}$
$=4\times 10^{- 3}$ .