Calculate the value of the dissociation constant for methanoic acid with a molar conductivity of 46.1 S cm² mol⁻¹ at 0.025 mol L⁻¹ concentration?
Given λ°(H⁺)= 349.6 S cm² mol⁻¹ and λ°(HCOO⁻) = 54.6 S cm² mol

1. \(1.27×10^{-4}~mol ~L^{−1}\)
2. \(5.17×10^{-5}~mol ~L^{−1}\)
3. \(3.67×10^{-4}~mol ~L^{−1}\)
4. \(4.87×10^{-5}~mol ~L^{−1}\)

The molar conductance of NaCl, HCI, and CH₃COONa at infinite dilution are 126.45, 426.16, and 91.0 S cm mol^–1 respectively. The molar conductance of CH₃COOH at infinite dilution will be:

The molar conductivity of 0.007 M acetic acid is 20 S cm² mol^–1. The dissociation constant of acetic acid is :

(\(\mathrm{\Lambda_{H^{+}}^{o} \ = \ 350 \ S \ cm^{2} \ mol^{-1} }\))
(\(\mathrm{\mathrm{\Lambda_{CH_{3}COO^{-}}^{o} \ = \ 50 \ S \ cm^{2} \ mol^{-1} }}\))

1. $1.75\times {10}^{-5}$ mol L^–1 

2. $2.50\times {10}^{-5}$ mol L^–1 

3. $1.75\times {10}^{-4}$ mol L^–1 

4. $2.50\times {10}^{-4}$ mol L^–1 

\(\Lambda _{m(NH_{4}OH)}^{o}\) is equal to :
1. \(\Lambda _{m(NH_{4}OH)}^{o} \ + \ \Lambda _{m(NH_{4}Cl)}^{o} \ - \ \Lambda _{m(HCl)}^{o}\)
2. \(\Lambda _{m(NH_{4}Cl)}^{o} \ + \ \Lambda _{m(NaOH)}^{o} \ - \ \Lambda _{m(NaCl)}^{o}\)
3. \(\Lambda _{m(NH_{4}Cl)}^{o} \ + \ \Lambda _{m(NaCl)}^{o} \ - \ \Lambda _{m(NaOH)}^{o}\)
4. \(\ \Lambda _{m(NaOH)}^{o} \ + \ \Lambda _{m(NaCl)}^{o}\ - \ \Lambda _{m(NH_{4}Cl)}^{o}\)

 ${\mathrm{\Lambda }}_{\mathrm{m}}^{°}{\mathrm{H}}_2\mathrm{O}$ can be represented by-

a.  ${{\Lambda }^{°}}_{m\left(\mathrm{HCl}\right)}+{{\Lambda }^{°}}_{m\left(\mathrm{NaOH}\right)}°-{\Lambda }_{m\left(\mathrm{NaCl}\right)}°$

b.  ${{\Lambda }^{°}}_{m\left({\mathrm{HNO}}_3\right)}+{{\Lambda }^{°}}_{m\left({\mathrm{NaNO}}_3\right)}-{\Lambda }_{m\left(\mathrm{NaOH}\right)}^{°}$

c.  ${{\Lambda }^{°}}_{m\left({\mathrm{HNO}}_3\right)}+{{\Lambda }^{°}}_{m\left(\mathrm{NaOH}\right)}-{\Lambda }_{m\left({\mathrm{NaNO}}_3\right)}^{°}$

d.  ${{\Lambda }^{°}}_{m\left({\mathrm{NH}}_4\mathrm{OH}\right)}+{{\Lambda }^{°}}_{\mathrm{m}\left(\mathrm{HCl}\right)}-{{\Lambda }^{°}}_{\mathrm{m}\left({\mathrm{NH}}_4\mathrm{Cl}\right)}$

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

Consider the following data:
Λ^°_m(Ca²⁺) = 119.0 S cm²mol^–1 
Λ^°_m(Cl^-) = 76.3 S cm²mol^–1
Λ^°_m(Mg²⁺) = 106.0 S cm²mol^–1
Λ^°_m(\(SO_{4}^{2-}\)) = 160.0 S cm²mol^–1 
 
The correct statement among the following is-

Λ^°_m for NaCl, HCl and NaAc are 126.4, 425.9 and 91.0 S cm²­­­­ mol^–1 respectively. The value of Λ^°_m for HAc is-

The conductivity of 0.001028 mol L^–1 acetic acid is 4.95 ×10^–5 S cm^–1.  if Λ^°_m for acetic acid is 390.5 S cm^2­­­­ mol^–1, Its dissociation constant value is-

1. 1.58 × 10^–5 mol L^–1
2. 1.78 × 10^–5 mol L^–1
3. 1.98 × 10^–5 mol L^–1
4. 2.18 × 10^–5 mol L^–1

The resistance of a cell containing 0.001 M KCl solution at 298 K is 1500 Ω. The conductivity is 0.146 × 10^–3 S cm^–1. The cell constant would be: