`^^_(m)^(0)H_(2)O` is equal to ___

Options

A. `^^_(m(HNO_(3)))^(0)+^^_(m(NaNO_(3)))^(0)-^^_(m(NaOH))^(0)`
B. `^^_(m(HCl))^(0)+^^_(m(NaOH))^(0)-^^_(m(NaCl))^(0)`
C. `^^_(m(HNO_(3)))^(0)+^^_(m(NaOH))^(0)-^^_(m(NaNO_(3)))^(0)`
D. `(^^_(m)^(0)(H_(2)SO_(4))-^^_(m)^(0)(K_(2)SO_(4)))/2+^^_(m(KOH))^(0)`

Explanation

To solve the question `^^_(m)^(0)H_(2)O`, we need to find the limiting molar conductivity (λ₀_m) of water (H₂O) using Kohlrausch's law. Let's break down the steps:

### Step 1: Understand Kohlrausch's Law
Kohlrausch's law states that the molar conductivity of an electrolyte at infinite dilution (λ₀_m) is equal to the sum of the molar conductivities of its individual ions.

### Step 2: Identify the Ions in Water
For water (H₂O), it can be dissociated into its ions:
- H₂O ⇌ H⁺ + OH⁻

Thus, the limiting molar conductivity of water can be expressed as:
\[ λ₀_m(H₂O) = λ₀_m(H⁺) + λ₀_m(OH⁻) \]

### Step 3: Write the Expression
Using the above relationship, we can write:
\[ λ₀_m(H₂O) = λ₀_m(H⁺) + λ₀_m(OH⁻) \]

### Step 4: Conclusion
The limiting molar conductivity of water is equal to the sum of the limiting molar conductivities of the hydrogen ion (H⁺) and the hydroxide ion (OH⁻).

Thus, the answer to the question `^^_(m)^(0)H_(2)O` is:
\[ λ₀_m(H₂O) = λ₀_m(H⁺) + λ₀_m(OH⁻) \]