The specific conductance of saturated solution os silver chloride is `k(ohm^(-1)cm^(-1))`. The limiting ionic conductance of `Ag^(+)` and `Cl^(-)` ions are x and y respectively. The solubility of AgCl in `"gram liter"^(-1)` is : (Molar mass of `AgCl=143.5"g mol"^(-1)`)

Options

A. `kxx(1000)/(x-y)`
B. `(k)/(x+y)xx143.5`
C. `(kxx1000xx143.5)/(x+y)`
D. `(x+y)/(k)xx(1000)/(143.5)`

Answer

To solve the problem, we need to find the solubility of silver chloride (AgCl) in grams per liter. We will use the given specific conductance (k) and the limiting ionic conductances (x for Ag⁺ and y for Cl⁻) to find the molar conductance and then relate it to solubility.

### Step-by-step Solution:

1. **Understand the relationship between molar conductance and limiting ionic conductances:**
The molar conductance (λ_m) of AgCl can be expressed as:
\[ \lambda_m = \lambda_{Ag^+} + \lambda_{Cl^-} = x + y \]

2. **Relate molar conductance to specific conductance:**
The molar conductance can also be related to specific conductance (κ) and molarity (C) as follows:
\[ \lambda_m = \frac{\kappa \times 1000}{C} \]
where κ is the specific conductance in ohm⁻¹ cm⁻¹ and C is the molarity in moles per liter.

3. **Rearranging the equation to find molarity:**
From the above equation, we can express molarity (C):
\[ C = \frac{\kappa \times 1000}{\lambda_m} \]
Substituting λ_m from step 1:
\[ C = \frac{\kappa \times 1000}{x + y} \]

4. **Relate molarity to solubility:**
The solubility (S) in grams per liter can be expressed in terms of molarity:
\[ S = C \times \text{molar mass} \]
Given that the molar mass of AgCl is 143.5 g/mol, we can substitute for C:
\[ S = \left(\frac{\kappa \times 1000}{x + y}\right) \times 143.5 \]

5. **Final expression for solubility:**
Thus, the solubility of AgCl in grams per liter is:
\[ S = \frac{\kappa \times 1000 \times 143.5}{x + y} \]

### Conclusion:
The solubility of AgCl in grams per liter is given by the formula:
\[ S = \frac{\kappa \times 1000 \times 143.5}{x + y} \]