The solubility product of AgCl is 1.8 × 10-10, the minimum volume ( in L ) of water required to dissolve 1.9 mg of AgCl is approximately.( molecular weight of AgCl =143.5 )

Question

The solubility product of AgCl is 1.8 × 10-10, the minimum volume ( in L ) of water required to dissolve 1.9 mg of AgCl is approximately.( molecular weight of AgCl =143.5 ) (A) 10 (B) 2 (C) 1 (D) 20

Tardigrade Admin · Accepted Answer

Solubility =√1.8 × 10-5 M =1.34 × 10-5 M =1.34 × 10-5 × 143.5 gm / litre =192.29 × 10-5 gm / litre =1.92 mg / litre so required volume =1 L

Q. The solubility product of $A g Cl$ is $1.8 \times 1 0^{- 10}$ , the minimum volume $($ in $L)$ of water required to dissolve $1.9 m g$ of $A g Cl$ is approximately.( molecular weight of $A g Cl = 143.5$ )

10

2

1

20

Solubility $= 1.8 \times 1 0^{- 5} M$
$= 1.34 \times 1 0^{- 5} M$
$= 1.34 \times 1 0^{- 5} \times 143.5 g m / l i t re$
$= 192.29 \times 1 0^{- 5} g m / l i t re$
$= 1.92 m g / l i t re$
so required volume $= 1 L$

Q. The solubility product of AgCl is 1.8×10−10, the minimum volume ( in L) of water required to dissolve 1.9mg of AgCl is approximately.( molecular weight of AgCl=143.5 )

10

2

1

20

Solubility =1.8​×10−5M =1.34×10−5M =1.34×10−5×143.5gm/litre =192.29×10−5gm/litre =1.92mg/litre so required volume =1L

Q. The solubility product of $A g Cl$ is $1.8 \times 1 0^{- 10}$ , the minimum volume $($ in $L)$ of water required to dissolve $1.9 m g$ of $A g Cl$ is approximately.( molecular weight of $A g Cl = 143.5$ )

Solubility $= 1.8 \times 1 0^{- 5} M$
$= 1.34 \times 1 0^{- 5} M$
$= 1.34 \times 1 0^{- 5} \times 143.5 g m / l i t re$
$= 192.29 \times 1 0^{- 5} g m / l i t re$
$= 1.92 m g / l i t re$
so required volume $= 1 L$