Limiting molar conductivities (in S cm^2 mol^-1) are: MgCl2 = 261.8, NaOH = 248.1, NaCl = 126.4. The limiting molar conductivity of Mg(OH)2 in S cm^2 mol^-1 (one decimal) is

JEE Main levelNumerical answerDifficulty 2 of 3ec194

Limiting molar conductivities (in $S c m^{2} mo l^{- 1}$ ) are: $MgCl X_{2} = 261.8$ , $NaOH = 248.1$ , $NaCl = 126.4$ . The limiting molar conductivity of $Mg (OH) X_{2}$ in $S c m^{2} mo l^{- 1}$ (one decimal) is

Numerical answer type, answer in S cm^2 mol^-1, tolerance 0.2.

Λ^{\circ} (Mg (OH) X_{2}) = λ^{\circ} (Mg X^{2 +}) + 2 λ^{\circ} (OH X^{-})

. From the given salts:

Λ^{\circ} (MgCl X_{2}) + 2 Λ^{\circ} (NaOH) - 2 Λ^{\circ} (NaCl)

removes two

Na X^{+}

and two

Cl X^{-}

= 261.8 + 2 (248.1) - 2 (126.4) = 261.8 + 496.2 - 252.8 = 505.2 S c m^{2} mo l^{- 1}