The Avogadro's number is $6 \times 10^{23} \ \text{mol}^{-1}$ and the electronic charge is $1.6 \times 10^{-19} \ \text{C}$. The Faraday's constant $(F)$ is given by:

  • A
    $6 \times 10^{23} \times 1.6 \times 10^{-19}$
  • B
    $\frac{6 \times 10^{23}}{1.6 \times 10^{-19}}$
  • C
    $\frac{2}{6 \times 10^{23} \times 1.6 \times 10^{-19}}$
  • D
    $\frac{1.6 \times 10^{-19}}{6 \times 10^{23}}$

Explore More

Similar Questions

$A$ solution of copper sulphate is electrolysed for $10 \text{ minutes}$ with a current of $1.5 \text{ amperes}$. The mass of copper deposited at the cathode is:
(Given: Molar mass of $Cu = 63 \text{ g mol}^{-1}$; $1F = 96487 \text{ C mol}^{-1}$)

$A$ current of $10.0 \, A$ flows for $2.00 \, h$ through an electrolytic cell containing a molten salt of metal $X$. This results in the decomposition of $0.250 \, mol$ of metal $X$ at the cathode. The oxidation state of $X$ in the molten salt is $(F = 96,500 \, C \, mol^{-1})$ (in $+$)

When the same quantity of electricity is passed through aqueous $H_2SO_4$ and molten $MgSO_4$,what is the ratio of the weights of hydrogen and magnesium deposited?

$A$ solution of $Fe_{2}(SO_{4})_{3}$ is electrolyzed for '$x$' min with a current of $1.5 \ A$ to deposit $0.3482 \ g$ of $Fe$. The value of $x$ is $.......$. [nearest integer]
Given : $1 \ F = 96500 \ C \ mol^{-1}$
Atomic mass of $Fe = 56 \ g \ mol^{-1}$

The number of electrons required to deposit $1 \ g$ atom of aluminium (at. wt. $= 27$) from a solution of aluminium chloride will be ............ $N$ (where $N$ is Avogadro's number).

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo