Advertisements
Advertisements
Question
\[\ce{Λ^0_m H2O}\] is equal to:
(i) \[\ce{Λ^0_m_{(HCl)} + \ce{Λ^0_m_{(NaOH)} - \ce{Λ^0_m_{(NaCl)}}}}\]
(ii) \[\ce{Λ^0_m_{(HNO_3)} + \ce{Λ^0_m_{(NaNO_3)} - \ce{Λ^0_m_{(NaOH)}}}}\]
(iii) \[\ce{Λ^0_{(HNO_3)} + \ce{Λ^0_m_{(NaOH)} - \ce{Λ^0_m_{(NaNO_3)}}}}\]
(iv) \[\ce{Λ^0_m_{(NH_4OH)} + \ce{Λ^0_m_{(HCl)} - \ce{Λ^0_m_{(NH_4Cl)}}}}\]
Advertisements
Solution
(i) \[\ce{Λ^0_m_{(HCl)} + \ce{Λ^0_m_{(NaOH)} - \ce{Λ^0_m_{(NaCl)}}}}\]
(iv) \[\ce{Λ^0_m_{(NH_4OH)} + \ce{Λ^0_m_{(HCl)} - \ce{Λ^0_m_{(NH_4Cl)}}}}\]
Explanation:
\[\ce{Λ^0_m_{(H_2O)} = \ce{Λ^0_m_{(HCl)} + \ce{Λ^0_m_{(NaOH)} - \ce{Λ^0_m_{(NaCl)}}}}}\]
\[\ce{Λ^0_{m(H^+)} + \ce{Λ^0_{m(OH^-)} = \ce{Λ^0_{m(H^+)} + \ce{Λ^0_{m(Cl^-)} + \ce{Λ^0_{m(Na^+)} + \ce{Λ^0_{m(OH^-)} - \ce{Λ^0_{m(Na^+)} - \ce{Λ^0_{m(Cl^-)}}}}}}}}}\]
\[\ce{Λ^0_m_{(HNO_3)} + \ce{Λ^0_m_{(NaOH)} - \ce{Λ^0_m_{(NaNO_3)} = \ce{Λ^0_m_{(H_2O)}}}}}\]
\[\ce{Λ^0_{m(H^+)} + \ce{Λ^0_{m(NO_3^-)} + \ce{Λ^0_{m(Na^+)} - \ce{Λ^0_{m(OH^-)} + \ce{Λ^0_{m(Na^+)} + \ce{Λ^0_{m(NO_3^-)} = \ce{Λ^0_{m(H^+)} - \ce{Λ^0_{m(OH^-)}}}}}}}}}\]
\[\ce{Λ^0_m_{(NH_4OH)} + \ce{Λ^0_m_{(HCl)} - \ce{Λ^0_m_{(NH_4Cl)} = \ce{Λ^0_m_{(H_2O)}}}}}\]
However, the sum of molar conductivities of constituent ions gives the molar conductivity of water but here \[\ce{NH4OH}\] is a weak electrolyte of which complete decomposition is not possible.
APPEARS IN
RELATED QUESTIONS
The molar conductivity of cation and anion of salt BA are 180 and 220 mhos respectively. The molar conductivity of salt BA at infinite dilution is_____________ .
(a) 90 mhos.cm2
(b) 110 mhos.cm2.mol-1
(c) 200 mhos.cm2.mol-1
(d) 400 mhos.cm2.mol-1
Define limiting molar conductivity.
The conductivity of 0.20 mol L−1 solution of KCl is 2.48 × 10−2 S cm−1. Calculate its molar conductivity and degree of dissociation (α). Given λ0 (K+) = 73.5 S cm2 mol−1 and λ0 (C1−) = 76.5 S cm2 mol−1.
Why does the conductivity of a solution decrease with dilution?
10.0 grams of caustic soda when dissolved in 250 cm3 of water, the resultant gram molarity of solution is _______.
(A) 0.25 M
(B) 0.5 M
(C) 1.0 M
(D) 0.1 M
How can you determine limiting molar conductivity, 0 m for strong electrolyte and weak electrolyte?
Which of the statements about solutions of electrolytes is not correct?
\[\ce{\Lambda^0_m(NH4OH)}\] is equal to ______.
Assertion: `"E"_("Ag"^+ //"Ag")` increases with increase in concentration of Ag+ ions.
Reason: `"E"_("Ag"^+ //"Ag")` has a positive value.
Assertion: Copper sulphate can be stored in zinc vessel.
Reason: Zinc is less reactive than copper.
Consider figure and answer the question to given below.
How will the concentration of Zn2+ ions and Ag+ ions be affected after the cell becomes ‘dead’?
The limiting molar conductivities Λ° for NaCl, KBr and KCl are 126, 152 and 150 S cm2 mol–1 respectively. The limiting molar conductivity Λ° for NaBr is ______.
Which of the following increases with the increase in the concentration of the solution?
Molar conductivity of substance “A” is 5.9 × 103 S/m and “B” is 1 × 10–16 S/m. Which of the two is most likely to be copper metal and why?
Conductivity of 2 × 10−3 M methanoic acid is 8 × 10−5 S cm−1. Calculate its molar conductivity and degree of dissociation if `∧_"m"^0` for methanoic acid, is 404 S cm2 mol−3.
The unit of molar conductivity is ______.
Which of the following solutions will have the highest conductivity at 298 K?
The resistance of a conductivity cell with a 0.1 M KCl solution is 200 ohm. When the same cell is filled with a 0.02 M NaCl solution, the resistance is 1100 ohm. If the conductivity of 0.1 M KCl solution is 0.0129 ohm-1 cm-1, calculate the cell constant and molar conductivity of 0.02 M NaCl solution.
The solution of two electrolytes A and B are diluted. ^m of B increases 1.5 times while that of A increases 25 times. Which of the two is a strong electrolyte? Give a reason.
