Limiting molar conductivity of \(H_2O\) is equal to:

\(\Lambda^o_m HCl + \Lambda^o_m NaOH − \Lambda^o_m NaCl\)

The limiting molar conductivity (\(\Lambda^o_m\)) of water (\(H_2O\)) is equal to the sum of the limiting molar conductivities of the ions it forms when it dissociates into ions. In the case of water, it forms \(H^+\) and \(OH^-\) ions when it dissociates. This can be done by evaluating the molar conductivities of \(HCl\), \(NaOH\) and \(NaCl\) as follows:

\(\Lambda^o_m HCl = \Lambda^o_m H^+ + \Lambda^o_mCl^-\) ------(1)

\(\Lambda^o_m NaOH = \Lambda^o_m Na^+ +\Lambda^o_m OH^-\) -------(2)

\(\Lambda^o_m NaCl = \Lambda^o_m Na^+ + \Lambda^o_m Cl^-\) ---------(3)

Now equation \(1 + 2 - 3\) is

 \(\Lambda^o_m HCl + \Lambda^o_m NaOH − \Lambda^o_m NaCl\)

\(= \Lambda^o_m H^+ + \Lambda^o_mCl^- + \Lambda^o_m Na^+ + \Lambda^o_m OH^- - \Lambda^o_m Na^+ - \Lambda^o_m Cl^-\)

\(= \Lambda^o_m H^+ + \Lambda^o_m OH^-\)

\(= \Lambda^o_m H_2O\)

So, the correct expression for the limiting molar conductivity of \(H_2O\) is: \(\Lambda^o_m H_2O = \Lambda^o_m H^+ + \Lambda^o_m OH^-\)

This can be achieved from option (2) \(\Lambda^o_m HCl + \Lambda^o_m NaOH − \Lambda^o_m NaCl\).