Find the incorrect equation from below:

(Λ_{m}^{0})_{N a B r} - (Λ_{m}^{0})_{N a C l} = (Λ_{m}^{0})_{K B r} - (Λ_{m}^{0})_{K C l}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(Λ_{m}^{0})_{H_{2} O} = (Λ_{m}^{0})_{H C l} + (Λ_{m}^{0})_{N a O H} - (Λ_{m}^{0})_{N a C l}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(Λ_{m}^{0})_{N a B r} - (Λ_{m}^{0})_{N a I} = (Λ_{m}^{0})_{K B r} - (Λ_{m}^{0})_{N a B r}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

(Λ_{m}^{0})_{K C l} - (Λ_{m}^{0})_{N a C l} = (Λ_{m}^{0})_{K B r} - (Λ_{m}^{0})_{N a B r}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $(Λ_{m}^{0})_{N a B r} - (Λ_{m}^{0})_{N a I} = (Λ_{m}^{0})_{K B r} - (Λ_{m}^{0})_{N a B r}$
Kohlrausch law of independent migration of ions:
$Λ_{m}^{0} = λ_{+}^{0} + λ_{-}^{0}$

$λ_{+}^{0} \to Limiting molar conductivities of cation$
$λ_{-}^{0} \to Limiting molar conductivities of anion$

In general,
$Λ_{m}^{0} = v_{+} λ_{+}^{0} + v_{-} λ_{-}^{0}$

$v_{+}$ and $v_{-}$ are the number of cations and anions after dissociation of an electrolyte.

In option (3),

$(Λ_{m}^{0})_{N a B r} - (Λ_{m}^{0})_{N a I} = (Λ_{m}^{0})_{B r^{+}} - (Λ_{m}^{0})_{I^{-}}$

$(Λ_{m}^{0})_{K B r} - (Λ_{m}^{0})_{N a B r} = (Λ_{m}^{0})_{K^{+}} - (Λ_{m}^{0})_{N a^{+}}$

Thus,
$(Λ_{m}^{0})_{N a B r} - (Λ_{m}^{0})_{N a I} \neq (Λ_{m}^{0})_{K B r} - (Λ_{m}^{0})_{N a B r}$

Hence, option (c) is correct.

Suggest Corrections