The specific conductance of a saturated solution of silver bromide is $k {Scm}^{- 1}$ . The limiting ionic conductivity of ${Ag}^{+}$ and ${Br}^{-}$ ions are $x$ and $y$ , respectively. The solubility of silver bromide in ${gL}^{- 1}$ is:

(molar mass of AgBr = 188)

Question

The specific conductance of a saturated solution of silver bromide is $k {Scm}^{- 1}$ . The limiting ionic conductivity of ${Ag}^{+}$ and ${Br}^{-}$ ions are $x$ and $y$ , respectively. The solubility of silver bromide in ${gL}^{- 1}$ is:

(molar mass of AgBr = 188)

Smart Achievers · Accepted Answer

(3)

$Λ_{m} = \frac{k}{M} \times 1000 \cdot (x + y) = \frac{k}{M} \times 1000$

$\Rightarrow M = \frac{1000 \times k}{x + y}$

$Solubility ({in g L}^{- 1}) = \frac{k \times 1000 \times 188}{x + y}$

Solution

Q.
The specific conductance of a saturated solution of silver bromide is $k {Scm}^{- 1}$ . The limiting ionic conductivity of ${Ag}^{+}$ and ${Br}^{-}$ ions are $x$ and $y$ , respectively. The solubility of silver bromide in ${gL}^{- 1}$ is:

(molar mass of AgBr = 188)

1 $\frac{k \times 1000}{x - y}$

2 $\frac{k}{x + y} \times 188$

3 $\frac{k \times 1000 \times 188}{x + y}$

4 $\frac{x + y}{k} \times \frac{1000}{188}$

Ans.
(3)

$Λ_{m} = \frac{k}{M} \times 1000 \cdot (x + y) = \frac{k}{M} \times 1000$

$\Rightarrow M = \frac{1000 \times k}{x + y}$

$Solubility ({in g L}^{- 1}) = \frac{k \times 1000 \times 188}{x + y}$

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Solution

Q. The specific conductance of a saturated solution of silver bromide is k Scm-1. The limiting ionic conductivity of Ag+ and Br- ions are x and y, respectively. The solubility of silver bromide in gL-1 is: (molar mass of AgBr = 188)

1 k×1000x-y

2 kx+y×188

3 k×1000×188x+y

4 x+yk×1000188

Ans. (3) Λm=kM×1000·(x+y)=kM×1000 ⇒ M=1000×kx+y Solubility (in g L-1)=k×1000×188x+y