The specific conductance $(K)$ of an electrolyte of $0.1 N$ concentration is related to equivalent conductance $(\land_{e})$ by the following formula.

\land_{e} = K

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\land_{e} = 10 K

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\land_{e} = 100 K

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\land_{e} = 10000 K

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is A $\land_{e} = 10000 K$
we have,
equivalent conductance = $\frac{K * 1000}{N}$

equivalent conductance= $\frac{K * 1000}{0.1 N}$

equivalent conductance= $10000 K$

Suggest Corrections

Similar questions

Q. The specific conductance (K) of an electrolyte of 0.1 N concentration is related to equivalent conductance (

Λ

) by the following

Q. (a) Following reactions occur at cathode during the electrolysis of aqueous silver chloride solution:

A g^{+} (a q) + e \to A g (s) E^{\circ} = + .80 V

H^{+} (a q) + e^{-} \to \frac{1}{2} H_{2} (g) E^{o} = 0.00 V

On the basis of their standard reduction electrode potential

(E^{o})

values, which reaction is feasible at the cathode and why?
(b) Define limiting molar conductivity. Why conductivity of an electrolyte solution decrease with decrease in concentration?

Q. Let ΔE denote the energy gap between the valence band and the conduction band. The population of conduction electrons (and of the holes) is roughly proportional to e⁻^Δ^E^/2kT. Find the ratio of the concentration of conduction electrons in diamond to the in silicon at room temperature 300 K. ΔE for silicon is 1.1 eV and for diamond is 6.1 eV. How many conduction electrons are likely to be in one cubic metre of diamond?

Q. Equivalent conductivity

\land_{e q}

The conductivity of a pure semiconductor is roughly proportional to $T^{3 / 2} e^{- Δ E / 2 K T}$ where $Δ E$ is the band gap. The band gap for germanium is 0.74 eV at 4 K and 0.67 eV at 300 K. By what factor does the conductivity of pure germanium increase as the temperature is raised from 4 K to 300 K ?

Join BYJU'S Learning Program

The specific conductance (K) of an electrolyte of 0.1 N concentration is related to equivalent conductance (∧e) by the following formula.

The correct option is A ∧e=10000Kwe have,equivalent conductance = K∗1000Nequivalent conductance= K∗10000.1Nequivalent conductance= 10000K

The specific conductance $(K)$ of an electrolyte of $0.1 N$ concentration is related to equivalent conductance $(\land_{e})$ by the following formula.

The correct option is A $\land_{e} = 10000 K$
we have,
equivalent conductance = $\frac{K * 1000}{N}$

equivalent conductance= $\frac{K * 1000}{0.1 N}$

equivalent conductance= $10000 K$