Advertisements
Advertisements
Question
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:
Options
100 Scm2/mol
115 Scm2/mol
150 Scm2/mol
125 Scm2/mol
Advertisements
Solution
115 Scm2/mol
Explanation:
Λ°CH3COOK = Λ°CH3COOH + Λ°KCl – Λ°HCl
= 390 + 150 – 425
= 115 Scm2/mol
APPEARS IN
RELATED QUESTIONS
Define limiting molar conductivity.
Why conductivity of an electrolyte solution decreases with the decrease in concentration ?
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?
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
Write mathematical expression of molar conductivity of the given solution at infinite dilution.
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?
Conductivity always decreases with decrease in concentration both, for weak and strong electrolytes because of the fact that ____________.
\[\ce{Λ^0_m H2O}\] 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) \[\ce{Λ^0_m_{(NH_4OH)} + \ce{Λ^0_m_{(HCl)} - \ce{Λ^0_m_{(NH_4Cl)}}}}\]
When acidulated water (dil.H2SO4 solution) is electrolysed, will the pH of the solution be affected? Justify your answer.
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’?
