Ncert Exercises & Solutions

$Λ_{m}^{°} H_{2} O$ can be represented by-

a. ${Λ^{°}}_{m (HCl)} + {Λ^{°}}_{m (NaOH)} ° - Λ_{m (NaCl)} °$

b. ${Λ^{°}}_{m ({HNO}_{3})} + {Λ^{°}}_{m ({NaNO}_{3})} - Λ_{m (NaOH)}^{°}$

c. ${Λ^{°}}_{m ({HNO}_{3})} + {Λ^{°}}_{m (NaOH)} - Λ_{m ({NaNO}_{3})}^{°}$

d. ${Λ^{°}}_{m ({NH}_{4} OH)} + {Λ^{°}}_{m (HCl)} - {Λ^{°}}_{m ({NH}_{4} Cl)}$

1. (a, b)
2. (b, c)
3. (c, d)
4. (a, c)

(a)

{Λ^{°}}_{m (HCl)} = {Λ^{°}}_{m (H^{+})} + {Λ^{°}}_{m ({Cl}^{-})}

{Λ^{°}}_{m (NaOH)} = {Λ^{°}}_{m ({Na}^{+})} + {Λ^{°}}_{m ({OH}^{-})}

{Λ^{°}}_{m (NaCl)} = {Λ^{°}}_{m ({Na}^{+})} + {Λ^{°}}_{m ({Cl}^{-})}

- -

{Λ^{°}}_{m (HCl)} + {Λ^{°}}_{m (NaOH)} - {Λ^{°}}_{m (NaCl)} = {Λ^{°}}_{m (H^{+})} + {Λ^{°}}_{m ({HO}^{-})}

(d)

{Λ^{°}}_{m ({NH}_{4} OH)} = Λ_{m ({NH}_{4}^{+})}^{°} + {Λ^{°}}_{m ({OH}^{-})}

{Λ^{°}}_{m (HCl)} = {Λ^{°}}_{m (H^{+})} + {Λ^{°}}_{m ({Cl}^{-})}

{Λ^{°}}_{m ({NH}_{4} Cl)} = {Λ^{°}}_{m ({NH}_{4}^{+})} + {Λ^{°}}_{m ({Cl}^{-})}

- - -

{Λ^{°}}_{m ({NH}_{4} OH)} + {Λ^{°}}_{m (HCl)} - {Λ^{°}}_{m ({NH}_{4} Cl)}

= Λ_{m (H^{+})} ° + Λ_{m ({OH}^{-})} ° = Λ_{m (H_{2} O)}^{0}

This type of decomposition is not possible due to weak basic strength of NH₄OH. This line will be placed above.

(b)

{Λ^{°}}_{m (H N O_{3})} = Λ °_{m (H^{+})} + Λ °_{m ({NO}_{3}^{-})}

Λ °_{m ({NaNO}_{3})} = Λ_{m ({Na}^{+})}^{0} + Λ_{m ({NO}_{3}^{-})}^{0}

{Λ^{°}}_{m (NaOH)} = Λ_{m ({Na}^{+})}^{°} + Λ °_{m (O H^{-})}

- - -

Λ °_{m ({HNO}_{3})} + Λ °_{m ({NaNO}_{3})} - Λ_{m (NaOH)}^{°}

= Λ_{m (H^{+})} ° + 2 Λ_{m ({NO}_{3}^{-})} - Λ_{m ({OH}^{-})} °

(b) is incorrect

(c)

{Λ^{°}}_{m (H N O_{3})} = Λ °_{m (H^{+})} + Λ °_{m ({NO}_{3}^{-})}

{Λ^{°}}_{m (NaOH)} = Λ_{m ({Na}^{+})}^{°} + Λ °_{m (O H^{-})}

Λ °_{m ({NaNO}_{3})} = Λ_{m ({Na}^{+})}^{0} + Λ_{m ({NO}_{3}^{-})}^{0}

- - -

Λ °_{m ({HNO}_{3})} + Λ_{m (NaOH)}^{°} - Λ °_{m ({NaNO}_{3})}

= Λ_{m (H^{+})} ° + 2 Λ_{m (O H^{-})} - Λ_{m (H_{2} O)} °

Hence, Options (a) and (c) are the correct choices.