Unit of Faraday is - Vidyadip

MCQ

Unit of Faraday is

A
Ampere
B
Coulomb
✓
Coulomb $mol{e^{ - 1}}$
D
Coulomb ${{\mathop{\rm Sec}\nolimits} ^{ - 1}}$

✓

Answer

Correct option: C.

Coulomb $mol{e^{ - 1}}$

c
In the International System of Units ( $SI$ ), the coulomb ($C$) is the preferred unit of electric charge quantity. It is equivalent to one ampere-second ($1\,A.s$) and represents approximately $6.24 \times 10^{18}$ electric charge carriers. To convert from coulombs to faradays, multiply by $1.04 \times 10^{-5}$.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

NEET STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

If the solubility product of $AB _{2}$ is $3.20 \times 10^{-11}\, M ^{3}$,

then the solubility of $AB _{2}$ in pure water is......... $\times 10^{-4}$ $mol$ $L ^{-1}$.

[Assuming that neither kind of ion reacts with water]

Which of the following reactions is an example of nucleophilic substitution reaction?

જ્યારે $C{H_2} = CH - COOH$ is reduced with $LiAl{H_4}$, the compound obtained will be

Which of the following is an electrophile

$K_{sp}$ for $BaSO_4$ at $25\,^oC$ is $1.1 \times 10^{-10}$. To make the new solubility equal to $1.1 \times 10^{-8}$, it is necessary to use the solution of $Na_2SO_4$ of the following concentration....$M$

In balmer series of lines of hydrogen spectrum, the third line from the red end corresponds to which one of the following inter-orbit jumps of the electron for Bohr orbits in an atom of hydrogen

The position of both an electron and a helium atom is known within $1.0\,nm$ and the momentum of the electron is known within $50 \times {10^{ - 26}}\,kg\,m{s^{ - 1}}$. The minimum uncertainty in the measurement of the momentum of the helium atom is

At a given temperature, the equilibrium constant for reaction $PC{l_5}_{(g)}$ $\rightleftharpoons$ $PC{l_3}_{(g)} + C{l_2}_{(g)}$ is $2.4 \times {10^{ - 3}}$. At the same temperature, the equilibrium constant for reaction $PC{l_3}_{(g)} + C{l_2}_{(g)}$ $\rightleftharpoons$ $PC{l_5}_{(g)}$ is

A solution containing $2.675\, g$ of $CoCl_3.6 NH_3$ (molar mass $= 267.5\, g\, mol^{-1}$) is passed through a cation exchanger. The chloride ions obtained in solution were treated with excess of $AgNO_3$ to give $4.78\, g$ of $AgCl$ (molar mass $= 143.5\, g\, mol^{-1}$). The formula of the complex is

(At. mass of $Ag= 108\,u$)

$CH_3 -CH_2 -CH=CH_2$ has hybridisation