The solubility of `[Co(NH)_(3))_(4)Cl_(4)]CIO_(4)`____if the `lambda_(Co(NH_(3))_(4)Cl_(2)^(+)=50, lambda_(ClO_(4)^(-)=70` and the measured resistance was `35.5 Omega` in a cell with cell constant of 0.20 is ____

Options

A. 59.75 mmol/L
B. 49.75 mmol/L
C. 39.75 mmol/L
D. 29.75 mmol/L

Explanation

To find the solubility of the complex \([Co(NH_3)_4Cl_2]ClO_4\), we will follow these steps:

### Step 1: Determine the Molar Conductance of the Complex
The complex dissociates into two ions:
1. \([Co(NH_3)_4Cl_2]^+\)
2. \(ClO_4^-\)

Given:
- Molar conductance of \([Co(NH_3)_4Cl_2]^+\) = 50 S cm²/mol
- Molar conductance of \(ClO_4^-\) = 70 S cm²/mol

The total molar conductance (\(\Lambda\)) of the complex can be calculated as:
\[ \Lambda = \Lambda_{[Co(NH_3)_4Cl_2]^+} + \Lambda_{ClO_4^-} = 50 + 70 = 120 \text{ S cm}^2/\text{mol} \]

### Step 2: Calculate the Conductance (K)
The relationship between conductance (K), resistance (R), and cell constant (X) is given by:
\[ K = \frac{X}{R} \]
Where:
- \(X = 0.20\) (cell constant)
- \(R = 35.5 \, \Omega\)

Calculating K:
\[ K = \frac{0.20}{35.5} = 0.005634 \text{ S/cm} \]

### Step 3: Relate Molar Conductance to Concentration
The formula for molar conductance is:
\[ \Lambda = \frac{1000 \times K}{C} \]
Where:
- \(\Lambda = 120 \text{ S cm}^2/\text{mol}\)
- \(K = 0.005634 \text{ S/cm}\)
- \(C\) is the concentration in mol/L.

Rearranging the formula to find \(C\):
\[ C = \frac{1000 \times K}{\Lambda} \]

### Step 4: Substitute Values to Find Concentration
Substituting the values:
\[ C = \frac{1000 \times 0.005634}{120} \]
Calculating \(C\):
\[ C = \frac{5.634}{120} = 0.04695 \text{ mol/L} \]

### Step 5: Convert to Millimoles
To convert the concentration from mol/L to millimol/L:
\[ C = 0.04695 \text{ mol/L} \times 1000 = 46.95 \text{ mmol/L} \]

### Final Answer
The solubility of \([Co(NH_3)_4Cl_2]ClO_4\) is approximately **46.95 mmol/L**.

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