20ml of AgNO3 solution was titrated with 0.20molL- 1 KCl solution in the conductivity cell. The data were plotted as follows. The moles of AgNO3 in the solution initially was

Question

20ml of AgNO3 solution was titrated with 0.20molL- 1 KCl solution in the conductivity cell. The data were plotted as follows. The moles of AgNO3 in the solution initially was <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/c-mhp6oajvpldzingo.jpg" /> (A) 2× 10- 3 (B) 4× 10- 3 (C) 8× 10- 3 (D) 2× 10- 2

Tardigrade Admin · Accepted Answer

At equivalence point Moles of textAgNO3=Moles of KCl 10- 3× 20× M=0.20× 20× 10- 3 M=0.20 Moles of AgNO3 initially =M× V =(0 . 20/100)× 20× 10- 3 =4× 10- 3 .

Q. $20 m l$ of $A g N O_{3}$ solution was titrated with $0.20 m o l L^{- 1}$ $K Cl$ solution in the conductivity cell. The data were plotted as follows. The moles of $A g N O_{3}$ in the solution initially was

$2 \times 1 0^{- 3}$

$4 \times 1 0^{- 3}$

$8 \times 1 0^{- 3}$

$2 \times 1 0^{- 2}$

At equivalence point
Moles of $AgNO_{3} = M o l es$ of $K Cl$
$1 0^{- 3} \times 20 \times M = 0.20 \times 20 \times 1 0^{- 3}$
$M = 0.20$
$M o l es$ of $A g N O_{3}$ initially $= M \times V$
$= \frac{0.20}{100} \times 20 \times 1 0^{- 3}$
$= 4 \times 1 0^{- 3}$ .

Q. 20ml of AgNO3​ solution was titrated with 0.20molL−1 KCl solution in the conductivity cell. The data were plotted as follows. The moles of AgNO3​ in the solution initially was

2×10−3

4×10−3

8×10−3

2×10−2