Advertisements
Advertisements
प्रश्न
Suggest a way to determine the \[\ce{\Lambda^{\circ}_m}\] value of water.
Advertisements
उत्तर
Water is a weak electrolyte. Its \[\ce{\Lambda^{\circ}_m}\] value can be determined with the help of Kohlrausch’s law.
\[\ce{\Lambda^{\circ}_m(HCl) = \Lambda^{\circ}_m(H+) + \Lambda^{\circ}_m(Cl^-)}\] ...(i)
\[\ce{\Lambda^{\circ}_m(NaOH) = \Lambda^{\circ}_m(Na+) + \Lambda^{\circ}_m(OH^-)}\] ...(ii)
\[\ce{\Lambda^{\circ}_m(NaCl) = \Lambda^{\circ}_m(Na+) + \Lambda^{\circ}_m(Cl^-)}\] ...(iii)
Adding (i) and (ii) and subtracting (iii), we get,
\[\ce{\Lambda^{\circ}_m(H2O) = \Lambda^{\circ}_m(H+) + \Lambda^{\circ}_m(OH^-)}\]
= \[\ce{\Lambda^{\circ}_m(HCl) + \Lambda^{\circ}_m(NaOH) - \Lambda^{\circ}_m(NaCl)}\]
APPEARS IN
संबंधित प्रश्न
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
The conductivity of 0.001 mol L-1 solution of CH3COOH is 3.905× 10-5 S cm-1. Calculate its molar conductivity and degree of dissociation (α) Given λ°(H+)= 349.6 S cm2 mol-1 and λ°(CH3COO)= 40.9S cm2mol-1.
The molar conductivity of 0.025 mol L−1 methanoic acid is 46.1 S cm2 mol−1. Calculate its degree of dissociation and dissociation constant. Given \[\ce{λ^0_{(H^+)}}\] = 349.6 S cm2 mol−1 and \[\ce{λ^0_{(HCOO^-)}}\] = 54.6 S cm2 mol−1.
The conductivity of 0.02 M AgNO3 at 25°C is 2.428 × 10−3 Ω−1 cm−1. What is its molar conductivity?
How can you determine limiting molar conductivity, 0 m for strong electrolyte and weak electrolyte?
In the plot of molar conductivity (∧m) vs square root of concentration (c1/2), following curves are obtained for two electrolytes A and B:
Answer the following:
(i) Predict the nature of electrolytes A and B.
(ii) What happens on extrapolation of ∧m to concentration approaching zero for electrolytes A and B?
\[\ce{Λ^0_m H2O}\] 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) \[\ce{Λ^0_m_{(NH_4OH)} + \ce{Λ^0_m_{(HCl)} - \ce{Λ^0_m_{(NH_4Cl)}}}}\]
Solutions of two electrolytes ‘A’ and ‘B’ are diluted. The Λm of ‘B’ increases 1.5 times while that of A increases 25 times. Which of the two is a strong electrolyte? Justify your answer.
Why on dilution the m Λm of \[\ce{CH3COOH}\] increases very fast, while that of \[\ce{CH3COONa}\] increases gradually?
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’?
Solutions of two electrolytes ‘A’ and ‘B’ are diluted. The Λm of ‘B’ increases 1.5 times while that of A increases 25 times. Which of the two is a strong electrolyte? Justify your answer. Graphically show the behavior of ‘A’ and ‘B’.
Which of the following halogen acids is the strongest reducing agent?
Which of the following increases with the increase in the concentration of the solution?
The molar conductance of NaCl, HCl and CH3COONa at infinite dilution are 126.45, 426.16 and 91.0 S cm2 mol−1 respectively. The molar conductance of CH3COOH at infinite dilution is ______.
Choose the right option for your answer.
Given below are two statements:
Statements I: The limiting molar conductivity of KCl (strong electrolyte) is higher compared to that of CH3COOH (weak electrolyte).
Statement II: Molar conductivity decreases with decrease in concentration of electrolyte.
In the light of the above statements, choose the most appropriate answer from the options given below:
Which of the following solutions will have the highest conductivity at 298 K?
Discuss the variation of conductivity and molar conductivity with concentration.
