A 100.0 mL dilute solution of Ag +is electrolysed for 15.0 minutes with a current of 1.25 mA and the silver is removed completely. What was the initial [ Ag +]?

Question

A 100.0 mL dilute solution of Ag +is electrolysed for 15.0 minutes with a current of 1.25 mA and the silver is removed completely. What was the initial [ Ag +]? (A) 2.32 × 10-1 (B) 2.32 × 10-4 (C) 2.32 × 10-3 (D) 1.16 × 10-5

Tardigrade Admin · Accepted Answer

No. of moles of A g+=(15 × 60+1.25 × 10-3/96,500) × (1/1) =0.0116 × 10-3 ∴ [A g+]=(1.16 × 10-5/(100)1000) =1.16 × 10-4

Q. A $100.0 m L$ dilute solution of $A g^{+}$ is electrolysed for $15.0$ minutes with a current of $1.25 m A$ and the silver is removed completely. What was the initial $[A g^{+}]$ ?

$2.32 \times 1 0^{- 1}$

$2.32 \times 1 0^{- 4}$

$2.32 \times 1 0^{- 3}$

$1.16 \times 1 0^{- 5}$

No. of moles of
$A g^{+} = \frac{15 \times 60 + 1.25 \times 1 0 ^{- 3}}{96 , 500} \times \frac{1}{1}$
$= 0.0116 \times 1 0^{- 3}$
$∴ [A g^{+}] = \frac{1.16 \times 1 0 ^{- 5}}{\frac{100}{1000}}$
$= 1.16 \times 1 0^{- 4}$

Q. A 100.0mL dilute solution of Ag+is electrolysed for 15.0 minutes with a current of 1.25mA and the silver is removed completely. What was the initial [Ag+]?

2.32×10−1

2.32×10−4

2.32×10−3

1.16×10−5