Calculate molar conductivity of `HCOOH` at infinite dilution, if equivalent conductivity of `H_(2)SO_(4)=x_(1), " " Al_(2)(SO_(4))_(3)=x_(2)," " (HCOO)_(3)Al=x_(3)`

Options

A. `6x_(1)-3x_(2)+6x_(3)`
B. `(x_(1)-x_(2)+x_(3))/(6)`
C. `x_(1)-x_(2)+x_(3)`
D. `(6x_(1)-3x_(2)+6x_(3))/(6)`

Answer

To calculate the molar conductivity of HCOOH at infinite dilution, we will use the equivalent conductivities of the given compounds and apply the relevant formulas. Here’s a step-by-step solution:

### Step 1: Understand the dissociation of the compounds
1. **H₂SO₄** dissociates into:
\[ H_2SO_4 \rightarrow 2H^+ + SO_4^{2-} \]
- Number of cations (H⁺) = 2
- Charge of cation (H⁺) = 1

2. **Al₂(SO₄)₃** dissociates into:
\[ Al_2(SO_4)_3 \rightarrow 2Al^{3+} + 3SO_4^{2-} \]
- Number of cations (Al³⁺) = 2
- Charge of cation (Al³⁺) = 3

3. **(HCOO)₃Al** dissociates into:
\[ (HCOO)_3Al \rightarrow 3HCOO^- + Al^{3+} \]
- Number of cations (Al³⁺) = 1
- Charge of cation (Al³⁺) = 3

### Step 2: Write the molar conductivity equations
Using the relationship between equivalent conductivity (Λ) and molar conductivity (Λm):
\[ \Lambda = \frac{n \cdot z \cdot \Lambda_m}{V} \]
Where:
- n = number of cations
- z = charge of cation

1. For H₂SO₄:
\[ \Lambda_{H_2SO_4} = 2 \cdot 1 \cdot \Lambda_m(H_2SO_4) \Rightarrow \Lambda_m(H_2SO_4) = \frac{\Lambda_{H_2SO_4}}{2} = \frac{x_1}{2} \]

2. For Al₂(SO₄)₃:
\[ \Lambda_{Al_2(SO_4)_3} = 2 \cdot 3 \cdot \Lambda_m(Al_2(SO_4)_3) \Rightarrow \Lambda_m(Al_2(SO_4)_3) = \frac{\Lambda_{Al_2(SO_4)_3}}{6} = \frac{x_2}{6} \]

3. For (HCOO)₃Al:
\[ \Lambda_{(HCOO)_3Al} = 1 \cdot 3 \cdot \Lambda_m((HCOO)_3Al) \Rightarrow \Lambda_m((HCOO)_3Al) = \frac{\Lambda_{(HCOO)_3Al}}{3} = \frac{x_3}{3} \]

### Step 3: Set up the equation for HCOOH
To find the molar conductivity of HCOOH, we will manipulate the equations derived from the dissociation of the compounds.

1. Multiply the equation for H₂SO₄ by 3:
\[ 3H_2SO_4 \rightarrow 6H^+ + 3SO_4^{2-} \]

2. Multiply the equation for (HCOO)₃Al by 2:
\[ 2(HCOO)_3Al \rightarrow 6HCOO^- + 2Al^{3+} \]

3. Add both equations:
\[ 3H_2SO_4 + 2(HCOO)_3Al \rightarrow 6H^+ + 6HCOO^- + 3SO_4^{2-} + 2Al^{3+} \]

4. Now, subtract the equation for Al₂(SO₄)₃:
\[ Al_2(SO_4)_3 \rightarrow 2Al^{3+} + 3SO_4^{2-} \]

### Step 4: Final equation and calculation
After adding and subtracting, we have:
\[ 3H_2SO_4 + 2(HCOO)_3Al - Al_2(SO_4)_3 \rightarrow 6H^+ + 6HCOO^- \]

Now, the molar conductivity of the resultant equation can be expressed as:
\[ \Lambda_m = 6 \cdot \Lambda_m(HCOOH) = 6 \cdot \left(\frac{x_1}{2} + \frac{x_3}{3} - \frac{x_2}{6}\right) \]

Dividing by 6 gives us:
\[ \Lambda_m(HCOOH) = \frac{1}{6}(6x_1 + 6x_3 - 6x_2) = x_1 + x_3 - x_2 \]

### Final Answer
The molar conductivity of HCOOH at infinite dilution is:
\[ \Lambda_m(HCOOH) = x_1 + x_3 - x_2 \]