The solubility of AgCl is 1.435 g/L What will be the solubility product?

Question

The solubility of AgCl is 1.435 g/L What will be the solubility product? (A)  10-4  (B) 0.01 (C) 0.001 (D) 0.1

Tardigrade Admin · Accepted Answer

undersets mathopAg Cl undersets mathopAg+ + undersets mathopCl- Solubility =1.435 g/L Solubility =(1.435/143.5)mol/L 143.5 = molecular weight Solubility product =(s)2 =(0.01)2=10-4

Q. The solubility of $A g Cl$ is 1.435 g/L What will be the solubility product?

$10^{- 4}$

0.01

0.001

0.1

$s A g Cl s A g^{+} + s C l^{-}$ Solubility =1.435 g/L Solubility $= \frac{1.435}{143.5} m o l / L$ 143.5 = molecular weight Solubility product $= (s)^{2}$ $= (0.01)^{2} = 10^{- 4}$

Q. The solubility of AgCl is 1.435 g/L What will be the solubility product?

10−4

0.01

0.001

0.1

sAg​ClsAg+​+sCl−​ Solubility =1.435 g/L Solubility =143.51.435​mol/L 143.5 = molecular weight Solubility product =(s)2 =(0.01)2=10−4

Q. The solubility of $A g Cl$ is 1.435 g/L What will be the solubility product?

$10^{- 4}$

$s A g Cl s A g^{+} + s C l^{-}$ Solubility =1.435 g/L Solubility $= \frac{1.435}{143.5} m o l / L$ 143.5 = molecular weight Solubility product $= (s)^{2}$ $= (0.01)^{2} = 10^{- 4}$