`wedge_(CICH_(2))^(infty) COONa =224 ohm ^(-1) cm^(2) "g equiv"^(-)`
`wedge_(NaCI)^(infty)=38.5 ohm ^(-1) "g equiv"^(-1)`
`wedge_(HCI)^(infty)=203 ohm ^(-1) cm^(2) "g equiv"^(-1)`
what is the value of `wedgeCICH_(2)COOH` ?

Options

A. `288.5Omega^(-1) cm^(2) g equiv ^(-1)`
B. `289.5Oemga^(-1) cm^(2) g equiv^(-1)`
C. ` 415.5 Omega^(-1) cm^(2)g equiv^(-1)`
D. `59.5 Omega^(-1) g equiv^(-1)`

Explanation

To find the value of the equivalent conductivity (λ) of the electrolyte \( \text{CICH}_2\text{COOH} \) at infinite dilution, we can use the principle of Kohlrausch's Law. According to this law, the equivalent conductivity of an electrolyte at infinite dilution is the sum of the conductivities of its constituent ions.

### Step-by-Step Solution:

1. **Identify the ions in the electrolyte**:
The electrolyte \( \text{CICH}_2\text{COOH} \) dissociates into \( \text{CICH}_2\text{COO}^- \) (the anion) and \( \text{H}^+ \) (the cation).

2. **Write the equation for the equivalent conductivity**:
According to Kohlrausch's Law:
\[ \lambda_{\text{CICH}_2\text{COOH}}^\infty = \lambda_{\text{CICH}_2\text{COO}^-}^\infty + \lambda_{\text{H}^+}^\infty \]

3. **Use the given data**:
We know:
- \( \lambda_{\text{NaCl}}^\infty = 38.5 \, \text{ohm}^{-1} \text{cm}^2 \text{g equiv}^{-1} \)
- \( \lambda_{\text{HCl}}^\infty = 203 \, \text{ohm}^{-1} \text{cm}^2 \text{g equiv}^{-1} \)
- \( \lambda_{\text{CICH}_2\text{COONa}}^\infty = 224 \, \text{ohm}^{-1} \text{cm}^2 \text{g equiv}^{-1} \)

4. **Set up the equations**:
From the data:
- For \( \text{NaCl} \):
\[ \lambda_{\text{Na}^+} + \lambda_{\text{Cl}^-} = 38.5 \]
- For \( \text{HCl} \):
\[ \lambda_{\text{H}^+} + \lambda_{\text{Cl}^-} = 203 \]
- For \( \text{CICH}_2\text{COONa} \):
\[ \lambda_{\text{CICH}_2\text{COO}^-} + \lambda_{\text{Na}^+} = 224 \]

5. **Solve the equations**:
From the first two equations, we can express \( \lambda_{\text{Cl}^-} \):
- From \( \lambda_{\text{Na}^+} + \lambda_{\text{Cl}^-} = 38.5 \):
\[ \lambda_{\text{Cl}^-} = 38.5 - \lambda_{\text{Na}^+} \]
- Substitute this into the second equation:
\[ \lambda_{\text{H}^+} + (38.5 - \lambda_{\text{Na}^+}) = 203 \]
Rearranging gives:
\[ \lambda_{\text{H}^+} = 203 - 38.5 + \lambda_{\text{Na}^+} = 164.5 + \lambda_{\text{Na}^+} \]

6. **Substituting back to find \( \lambda_{\text{CICH}_2\text{COO}^-} \)**:
Substitute \( \lambda_{\text{H}^+} \) back into the equation for \( \text{CICH}_2\text{COONa} \):
\[ \lambda_{\text{CICH}_2\text{COO}^-} + \lambda_{\text{Na}^+} = 224 \]
We can express \( \lambda_{\text{CICH}_2\text{COO}^-} \) in terms of \( \lambda_{\text{Na}^+} \):
\[ \lambda_{\text{CICH}_2\text{COO}^-} = 224 - \lambda_{\text{Na}^+} \]

7. **Final calculation**:
Now, substituting \( \lambda_{\text{H}^+} \) into the equation for \( \lambda_{\text{CICH}_2\text{COOH}} \):
\[ \lambda_{\text{CICH}_2\text{COOH}}^\infty = \lambda_{\text{CICH}_2\text{COO}^-} + \lambda_{\text{H}^+} \]
Substitute \( \lambda_{\text{CICH}_2\text{COO}^-} \) and \( \lambda_{\text{H}^+} \):
\[ \lambda_{\text{CICH}_2\text{COOH}}^\infty = (224 - \lambda_{\text{Na}^+}) + (164.5 + \lambda_{\text{Na}^+}) = 224 + 164.5 = 388.5 \, \text{ohm}^{-1} \text{cm}^2 \text{g equiv}^{-1} \]

### Final Answer:
\[ \lambda_{\text{CICH}_2\text{COOH}}^\infty = 388.5 \, \text{ohm}^{-1} \text{cm}^2 \text{g equiv}^{-1} \]