The specific conductance of a saturated solution of AgCl at 25 °C is 1.821 × 10-5 mho cm-1. What is the solubility of AgCl in water (in g L-1), if limiting molar conductivity of AgCl is 130.26 mho cm2 mol-1 ?

Question

The specific conductance of a saturated solution of AgCl at 25 °C is 1.821 × 10-5 mho cm-1. What is the solubility of AgCl in water (in g L-1), if limiting molar conductivity of AgCl is 130.26 mho cm2 mol-1 ? (A) 1.89 × 10-3 g L-1 (B) 2.78 × 10-2 g L-1 (C) 2.004×10-2 g L-1 (D) 1.43 × 10-3 g L-1

Tardigrade Admin · Accepted Answer

Solubility =( kappa×1000/ Lambda°m) =(1.821×10-5×1000/130.26) =13.97×10-5 mol L-1 =13.97×10-5×143.5 (AgCl=108+35.5=143.5) =2.004×10-2 g L-1

Q. The specific conductance of a saturated solution of $A g Cl$ at $25^{\circ} C$ is $1.821 \times 1 0^{- 5} mh o c m^{- 1}$ . What is the solubility of $A g Cl$ in water (in $g L^{- 1}$ ), if limiting molar conductivity of $A g Cl$ is $130.26 mh o c m^{2} m o l^{- 1}$ ?

$1.89 \times 1 0^{- 3} g L^{- 1}$

$2.78 \times 1 0^{- 2} g L^{- 1}$

$2.004 \times 1 0^{- 2} g L^{- 1}$

$1.43 \times 1 0^{- 3} g L^{- 1}$

Solubility $= \frac{κ \times 1000}{Λ _{m}^{\circ}}$
$= \frac{1.821 \times 1 0 ^{- 5} \times 1000}{130.26}$
$= 13.97 \times 1 0^{- 5} m o l L^{- 1}$
$= 13.97 \times 1 0^{- 5} \times 143.5 (A g Cl = 108 + 35.5 = 143.5)$
$= 2.004 \times 1 0^{- 2} g L^{- 1}$

Q. The specific conductance of a saturated solution of AgCl at 25∘C is 1.821×10−5mhocm−1. What is the solubility of AgCl in water (in gL−1), if limiting molar conductivity of AgCl is 130.26mhocm2mol−1 ?

1.89×10−3gL−1

2.78×10−2gL−1

2.004×10−2gL−1

1.43×10−3gL−1