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
Define “Molar conductivity”.
Resistance of conductivity cell filled with 0.1 M KCl solution is 100 ohms. If the resistance of the same cell when filled with 0.02 M KCl solution is 520 ohms, calculate the conductivity and molar conductivity of 0.02 M KCl solution. [Given: Conductivity of 0.1 M KCl solution is 1.29 S m-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.
Define the following terms: Molar conductivity (⋀m)
Conductivity of 0.00241 M acetic acid is 7.896 × 10−5 S cm−1. Calculate its molar conductivity and if `Lambda_m^0` for acetic acid is 390.5 S cm2 mol−1, what is its dissociation constant?
The conductivity of 0.02 M AgNO3 at 25°C is 2.428 × 10−3 Ω−1 cm−1. What is its molar conductivity?
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?
\[\ce{\Lambda^0_m(NH4OH)}\] is equal to ______.
Match the items of Column I and Column II on the basis of data given below:
`E_("F"_2//"F"^-)^Θ` = 2.87 V, `"E"_(("Li"^(+))//("Li"^-))^Θ` = − 3.5V, `"E"_(("Au"^(3+))//("Au"))^Θ` = 1.4 V, `"E"_(("Br"_(2))//("Br"^-))^Θ` = 1.09 V
| Column I | Column II |
| (i) F2 | (a) metal is the strongest reducing agent |
| (ii) Li | (b) metal ion which is the weakest oxidising agent |
| (iii) Au3+ | (c) non metal which is the best oxidising agent |
| (iv) Br– | (d) unreactive metal |
| (v) Au | (e) anion that can be oxidised by Au3+ |
| (vi) Li+ | (f) anion which is the weakest reducing agent |
| (vii) F– | (g) metal ion which is an oxidising agent |
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’?
An increase in equivalent conductance of a strong electrolyte with dilution is mainly due to :-
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.
The molar conductivity of CH3COOH at infinite dilution is 390 Scm2/mol. Using the graph and given information, the molar conductivity of CH3COOK will be:

Which of the following solutions of KCl will have the highest value of molar conductivity?
Assertion (A) : Conductivity decreases with decrease in concentration of electrolyte.
Reason (R) : Number of ions per unit volume that carry the current in a solution decreases on dilution.
The specific conductance of 2.5 × 10-4 M formic acid is 5.25 × 10-5 ohm-1 cm-1. Calculate its molar conductivity and degree of dissociation.
Given `λ°_("H"^+)` = 349.5 ohm-1 cm2 mol-1 and
`λ°_("HCOO"^-) = 50.5 " ohm"^-1 "cm"^2 "mol"^-1`
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.
